Chapter 2 Electric Charges and Fields (Electrostatics Part 2) – Physics free study material by TEACHING CARE online tuition and coaching classes

Chapter 2 Electric Charges and Fields (Electrostatics Part 2) – Physics free study material by TEACHING CARE online tuition and coaching classes

File name : Chapter-2-Electric-Charges-and-Fields-Electrostatics-Part-2.pdf

 

(ii) Charged circular ring : Suppose we have a charged circular ring of radius R and charge Q. On it’s axis electric field and potential is to be determined, at a point ‘x’ away from the centre of the ring.

 

 

  • Electric field : Consider an element carrying charge dQ . It’s electric field

dE = (

KdQ

) directed

 

R 2 + x 2

as shown. It’s component along the axis is

dE cosq

and perpendicular to the axis is

dE sinq . By

 

symmetry

dE sinq

= 0 , hence E =

dE cosq =

kdQ       .       x

 

 

ò                          ò            ò (R 2 + x 2 )

 

E =      kQx    

(R 2  + x 2 )3  2

directed away from the centre if Q is positive

 

(b)  Potential :

V =     1   .      Q

4pe 0

 

Note : At centre x = 0 so Ecentre= 0 and

Vcentre

kQ

R

 

□ At a point on the axis such that x >> R

E kQ

Q

x 2

and

V kQ

x

 

□ At a point on the axis if

(3)  Surface charge :

x = ±  R  ,

Emax =

 

  • Infinite sheet of charge : Electric field and potential at a point P as shown

 

E =   s

2e 0

0

s r

(E µ r o )

 

and

V = – 2e    + C

 

  • Electric field due to two parallel plane sheet of charge : Consider two large, uniformly charged Plates A and B, having surface charge

 

densities are s A

and s B

respectively. Suppose net electric

 

field at points P, Q and R is to be calculated.

 

 

At P,

 

At Q,

EP    = (EA

 

E     = (E

  • EB ) =

 

E ) =

1

2e 0

A
B

1

(s A

 

A

(s

+ s B

 

B

s

)

 

);   At R, E

 

 

= -(E

 

 

+ E ) = –

 

1 (s

 

 

+ s )

 

Q                  A               B

2e 0

R                      A                B

2e 0

 

 

 

Note :         If

s A = +s

and

s B = –s

then

Ep  = 0, EQ

= s , E

R

e 0

 

= 0 . Thus in case of two infinite

 

plane sheets of charges having equal and opposite surface charge densities, the field is non-zero only in the space between the two sheets and is independent of the distance between them i.e., field is uniform in this region. It should be noted that this result will hold good for finite plane sheet also, if they are held at a distance much smaller then the dimensions of sheets i.e., parallel plate capacitor.

(iii)  Conducting sheet of charge :

E = s

e 0

sr

e

V = –      + C

0

 

 

  • Charged conducting sphere : If charge on a conducting sphere of radius R is Q as shown in figure then electric field and potential in different situation are –

(a)   Out side the sphere : P is a point outside the sphere at a distance r from the centre at which electric field and potential is to be determined.

Electric field at P

 

E          1  . Q

sR 2

and V

   1   . QsR 2

ìQ s ´ A

 

out

4pe 0 r 2

e 0r 2

out

4pe 0    r

e 0r      î

=s ´ 4pR 2

 

í
  • At the surface of sphere : At surface r = R

 

 

So,

E      1   . Q  s

and V

   1   . QsR

 

0
0
0
0

s       4pe     R 2    e              s      4pe     R       e

(c)   Inside the sphere : Inside the conducting charge sphere electric field is zero and potential remains constant every where and equals to the potential at the surface.

 

Ein = 0

and

Vin = constant = Vs

 

 

 

 

Note : @Graphical variation of electric field and potential of a charged spherical conductor with distance

 

 

 

 

Er graph

V r graph

 

 

 

 

 

 

 

 

r                                                                                                                                            r

 

 

(4)  Volume charge (charged non-conducting sphere) :

Charge given to a non conducting spheres spreads uniformly throughout it’s volume.

(i)  Outside the sphere at P

 

Eout

=     1

4pe 0

. Q and

r 2

Vout

=     1   . Q

4pe 0    r

by using

r =     Q

4 pR 3

3

 

 

Eout

=   rR 3     and

3e 0r 2

Vout

rR 3

3e 0r

 

  • At the surface of sphere : At surface r = R

 

E      1   . Q  =  rR

0

and

V   =     1   . Q =

 

rR 2

 

 

0

s       4pe     R 2    3e

0

s       4pe     R

3e 0

 

  • Inside the sphere : At a distance r from the centre

 

E        1  . Qr

= rr    {E

µ r}

0

and

V    =    1

Q[3R 2r 2 ] = r(3R 2r 2 )

 

0

in       4pe     R 3

3e 0

in      4pe

2R 3

6e 0

 

in

Note : @            At centre

r = 0

So,

V          = 3 ´     . Q = 3 V

i.e.,

 

 

s

Vcentre

  • Vsurface
  • Vout

centre

2   4pe 0 R       2

 

@          Graphical variation of electric field and potential with distance

 

 

 

E-r graph

V-r graph

 

 

 

 

 

 

 

 

 

r                                                                                                                                                       r

r = R

 

 

 

(5)  Electric field and potential in some other cases

  • Uniformly charged semicircular ring : l = charge

length

At centre :

E 2Kl =           Q        

 

 

 

 

 

+Q

+     +    +

+                      +

+                              +

 

 

+                                        +

R          2p 2e 0 R 2

R

V KQ =     Q       

R         4pe 0 R

 

(iii) Charged cylinder of infinite length

(a) Conducting                                                   (b) Non-conducting

 

 

P

r

 

 

 

 

R

+  +           +  +

+  +           +  +

+  +           +  +              P

+

+   +          +   r

+  +           +  +

 

 

 

 

 

For both type of cylindrical charge distribution E

=     l    , and E

    l   

but for conducting

 

out

2pe 0r

suface

2pe 0 R

 

 

E     = 0 and for non-conducting E

=      lr      . (we can also write formulae in form of r i.e., E

rR 2

etc.)

 

0
0

in                                                                                           in       2pe R 2                                                                                             out       2e r

 

 

(ii) Hemispherical charged body :

s

 

+ + + +

 

At centre O,   E =                                                                          +      + +     +

+         +

 

4e 0

+  +   + +

+ + +

 

 

V sR

2e 0

+  + O · + +  +

 

(iv) Uniformly charged disc

At a distance x from centre O on it’s axis

 

= s   é

         x         ù                                                      s

 

E      2e

ê1 –

0 êë

x 2 + R

ú

2 úû                                                   R

O

 

ê

V   s   é

2e 0 ë

x 2 + R 2

x

ú

xù

û

 

Note :        @    Total charge on disc Q = spR2

 

@    If x ® 0,

E ~–   s

2e 0

i.e. for points situated near the disc, it behaves as an infinite sheet of charge.

 

 

 

Concepts

No point charge produces electric field at it’s own position.

Since charge given to a conductor resides on it’s surface hence electric field inside it is zero.

++ +                                                                    +

 

+       +  +

+   +                         +

+

E = 0             +

+

+

+            +          +

+ +        +  +

+                       +

+         E = 0         +

+                         +

+                   +

+   +   +

 

The electric field on the surface of a conductor is directly proportional to the surface charge density at that point i.e, E µ s

Two charged spheres having radii r1 and r2 charge densities s 1 and s 2 respectively, then the ratio of electric field on their

 E1      s 1     r 2              ì          Q

 

surfaces will be

=        = 2

ís =        2

 

r
1

E2      s 2      2

î        4pr

 

In air if intensity of electric field exceeds the value 3 ´ 106 N/C

air ionizes.

 

A small ball is suspended in a uniform electric field with the help of an insulated thread. If a high energy x–ray bean falls on the ball, x-rays knock out electrons from the ball so the ball is positively charged and therefore the ball is deflected in the direction of electric field.

r

E

F= QE

X–Ray

 

Electric field is always directed from higher potential to lower potential.

A positive charge if left free in electric field always moves from higher potential to lower potential while a negative charge moves from lower potential to higher potential.

The practical zero of electric potential is taken as the potential of earth and theoretical zero is taken at infinity. An electric potential exists at a point in a region where the electric field is zero and it’s vice versa.

A point charge +Q lying inside a closed conducting shell does not exert force another point charge q placed outside the shell as

 

shown in figure

+   +

+Q +               +

+       –  –   –       +                        +q

 

+       –  +Q  –       +

+       –  –         +

+

+                +

+

Actually the point charge +Q is unable to exert force on the charge +q because it can not produce electric field at the position of

+q. All the field lines emerging from the point charge +Q terminate inside as these lines cannot penetrate the conducting medium (properties of lines of force).

The charge q however experiences a force not because of charge +Q but due to charge induced on the outer surface of the shell.

 

 

 

æ            1   ö

Example: 21      A half ring of radius R has a charge of l per unit length. The electric field at the centre is ç k =                              ÷

è        4pe0 ø

[CPMT 2000; CBSE PMT 2000; REE 1999]

 

 

 

Solution: (c)

  • Zero (b)

 

dl = Rdq

 

 

Charge on dl = lRdq .

kl                            (c)

R

2kl

 

R

dl

dq

q

q

(d)

 

 

 

C

q

kpl R

 

Field at C due to dl = k lRdq . = dE

R2

We need to consider only the component

dl¢

 

dE cos q , as the component

 

dE sin q

dE

 

will cancel out because

 

of the field at C due to the symmetrical element dl¢,

 

ò

p 2

The total field at C is = 2        dE cos q

= 2 kl òp  2 cosqdq  = 2k  l

ìï=       Q       üï

 

0                               R   0

í

ï
þ

R           î  2pe 0 R

2 ýï

 

Example: 22      What is the magnitude of a point charge due to which the electric field 30 cm away has the magnitude 2

 

newton/coulomb[1 / 4pe 0 = 9 ´ 109 Nm2 ]

[MP PMT 1996]

 

(a)

2 ´ 10 11 coulomb

9 ´ 10 11 coulomb

(b)

3 ´ 10 11 coulomb

(c)

5 ´ 10 11 coulomb

(d)

 

Solution: (a)     By using

E =     1

4pe0

. Q ;

r 2

2 = 9 ´ 109 ´

Q            Þ Q = 2 ´ 1011C

(         2)

30 ´ 10-2

 

Example: 23      Two point charges Q and – 3Q are placed at some distance apart. If the electric field at the location of Q

is E, then at the locality of – 3Q, it is

 

  • E
  • E/3 (c)

-3E

  • E/3

 

 

 

Solution: (b)     Let the charge Q and – 3Q be placed respectively at A and B at a distance x

Now we will determine the magnitude and direction to the field produced by charge – 3Q at A, this is E

as mentioned in the Example.                                                         A                                                                               B

 

\ E = 3Q

x2

(along AB directed towards negative charge)

Q                                                                                                   – 3 Q

x

 

Now field at location of – 3Q i.e. field at B due to charge Q will be away from positive charge)

E’ =  Q  = E

x2      3

(along AB directed

 

Example: 24      Two charged spheres of radius

R1 and

R2 respectively are charged and joined by a wire. The ratio of

 

electric field of the spheres is                                                                        [CPMT 1999 Similar to CBSE 2003]

 

(a)

 R1

R2

(b)

 R2

R1

2

R
R
2
  •  1 2

2

R
R
2
  • 2 1

 

Solution: (b)     After connection their potential becomes equal i.e., k . Q1

R1

k . Q2 ; Þ

R2

Q1 = R1 Q2           R2

 

 

E         Q        æ R   ö 2      R

 

Ratio of electric field  1 =  1 ´ ç  2 ÷

=    2 .

 

E2        Q2

è R1 ø          R1

 

Example: 25      The number of electrons to be put on a spherical conductor of radius 0.1m to produce an electric field of

0.036 N/C just above its surface is                                                                                                 [MNR 1994]

 

(a)

2.7 ´ 105

(b)

2.6 ´ 105

(c)

2.5 ´ 105

(d)

2.4 ´ 105

 

 

Solution: (c)      By using

E = k

Q , where R = radius of sphere so 0.036 = 9 ´ 109 ´

R2

ne

 

(0.1)2

Þ n = 2.5 ´ 105

 

Example: 26      Eight equal charges each +Q are kept at the corners of a cube. Net electric field at the centre will be

æ           1   ö

çk =            ÷

è        4pe0 ø

 

 

(a)

kQ                                          (b)

r 2

8kQ                                   (c)

r 2

2kQ                               (d) Zero

r 2

 

Solution: (d)     Due to the symmetry of charge. Net Electric field at centre is zero.

 

+q

Note : q

q                          a                       q

 

 

a                                                    a

 

 

 

 

+q                                                           +q

 

Equilateral triangle

q                                   q

Square

 

 

Example: 27      q, 2q, 3q and 4q charges are placed at the four corners A, B, C and D of a square. The field at the centre

O of the square has the direction along.

[CPMT 1989]

 

 

 

 

 

 

 

 

 

 

 

(a) AB                                           (b) CB                                       (c)  AC                                   (d) BD

Solution: (b)     By making the direction of electric field due to all charges at centre. Net electric field has the direction along

CB

Example: 28      Equal charges Q are placed at the vertices A and B of an equilateral triangle ABC of side a. The magnitude of electric field at the point A is

 

 

(a)

Q

 

4pe0a2

(b)

4pe0a2

(c)

4pe0a2

(d)

Q

 

2pe0a2

 

Solution: (c)     As shown in figure Net electric field at A

 

 

 

E =

1      Q

0

EB = EC = 4pe   . a2

 

 

 

So,

E     3Q

4pe0a2

+ B                 a                C +

 

Example: 29      Four charges are placed on corners of a square as shown in figure having side of 5 cm. If Q is one micro coulomb, then electric field intensity at centre will be                                                                         [RPET 1999]

 

(a)

(b)

(c)

(d)

1.02 ´ 107 N / C upwards

2.04 ´ 107 N / C downwards

2.04 ´ 107 N / C upwards

1.02 ´ 107 N / C downwards

Q                                             – 2Q

 

 

 

Q                                            + 2Q

 

Solution: (a)

| EC| > | E A| so resultant of EC & EA  is

ECA = EC EA directed toward Q              Q

– 2Q

 

Also | E

A                                                       B

| > | E |so resultant of E   and E       i.e.

 

B                      D                                                   B                    D

EBD = EB ED directed toward – 2Q charge hence Net electric field at centre is

 

E =                                                                                                             .… (i)

 

 

By proper calculations | EA |= 9 ´ 109 ´

æ

10 -6

5

D

Q

= 0.72 ´ 107 N/C

ö 2

C

+2Q

 

ç        ´ 10 -2 ÷

è                     ø

 

 

 

| EB

|= 9 ´ 109 ´

æ

2 ´ 10 -6

5            ö 2

= 1.44 ´ 107 N/C ;

| EC

|= 9 ´ 109 ´

æ

2 ´ 10 -6

5            ö 2

= 1.44 ´ 107 N/C

 

ç        ´ 10 -2 ÷

è                     ø

ç        ´ 10 -2 ÷

è                     ø

 

 

| E   |= 9 ´ 109 ´

10 -6

= 0.72 ´ 107 N/C ;     So, |E

|= |E ||E  |= 0.72 ´ 107 N/C

 

 

5

D                          æ                     ö 2

ç        ´ 10 -2 ÷

è                     ø

CA               C                A

 

and | EBD |=| EB |-| ED |= 0.72 ´ 107 N/C. Hence from equation – (i) E = 1.02 ´ 107 N/C upwards

Example: 30      Infinite charges are lying at x = 1, 2, 4, 8…meter on X-axis and the value of each charge is Q. The value of intensity of electric field and potential at point x = 0 due to these charges will be respectively

(a) 12 ´ 109 Q N/C, 1.8 ´ 104 V                                                 (b) Zero, 1.2 ´ 104V

 

 

(c)

6 ´ 109 Q N/C, 9 ´ 103 V                                                       (d)

4 ´ 109 Q

N/C , 6 ´ 103 V

 

Solution: (a)     By the superposition, Net electric field at origin

E = kQé 1 + 1 + 1 + 1 + …¥ù

êë12      22      4 2      82            úû

 

 

E = kQé1 + 1 + 1 + 1 + …¥ù

ëê     4    16    64         úû

 

1 + 1 + 1 + 1

4    16    64

+ …¥

is an infinite geometrical progression it’s sum can be obtained by using the

 

¥

formula S  =   a 

1 – r

; Where a = First term, r = Common ratio.

 

Here a = 1 and r = 1

4

so, 1 + 1 + 1 + 1

4    16    64

+ ….. ¥ =

1

 

1 – 1 / 4

= 4 .

3

 

Hence E = 9 ´ 109 ´ Q ´ 4 = 12 ´ 109 Q N/C

3

ë
ú
û

Electric potential at origin V =  1  é1 ´ 10-6 + 1 ´ 10-6 + 1 ´ 10-6 + 1 ´ 10-6 +…………………. ¥ù

 

4pe 0 ê        1

2             4              8                 ú

é           ù

 

= 9 ´ 109 ´ 10 -6 é1 + 1 + 1 + 1 + …………¥ù = 9 ´ 103 ê 1 ú = 1.8 ´ 104 volt

 

1

ëê     2    4    8

ú                   ê

ê

û                   ê1 –

ë

ú

ú

2 úû

 

Note : @                                       In the arrangement shown in figure +Q and – Q are alternatively and

equally spaced from each other, the net potential at the origin O is V =  1 . Q log e 2

4pe 0         x

 

 

+Q             Q             +Q             Q                                      

O                      x                 2x               3x               4x                                      

 

Example: 31 Potential at a point x-distance from the centre inside the conducting sphere of radius R and charged with charge Q is                                                                                                                        [MP PMT 2001]

 

(a)

Q                                            (b)

R

Q                                        (c)

x

Q                                   (d) xQ

x 2

 

Solution: (a)     Potential inside the conductor is constant.

Example: 32      The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9 ´ 105 V is

 

  • 80 V (b)

8 ´ 106 V

  • 9 V (d)

9 ´ 105 V

 

 

Solution: (b)

V = 9 ´ 109 ´ ne = 9 ´ 109 ´ 50 ´ 1.6 ´ 10 -19

= 8 ´ 106 V

 

r                                9 ´ 10 -15

Example: 33      Eight charges having the valves as shown are arranged symmetrically on a circle of radius 0.4m in air.

Potential at centre O will be

 

 

 

(a)

63 ´ 104 volt

(b)

63 ´ 1010 volt

(c)

63 ´ 106 volt

  • Zero

 

Solution: (a)     Due to the principle of superposition potential at O

 

V    1   ´ 28 ´ 10 -6

= 9 ´ 109 ´ 28 ´ 10 -6

= 63 ´ 104 volt

 

4pe 0          0.4

0.4

 

Example: 34      As shown in the figure, charges +q and –q are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is                                                                                                                           [MP PET 2000]

 

 

 

 

 

 

 

 

 

(a)

1    .      2q

4pe 0

(b)

1    .       q

4pe 0

(c)

1    .

4pe 0

(-q)

  • Zero

 

Solution: (d)     Potential at A = Potential due to (+q) charge + Potential due to (– q) charge

 

=      1    .       q          +     1

4pe 0

(-q)     = 0

 

Example: 35      A conducting sphere of radius R is given a charge Q. consider three points B at the surface, A at centre and C at a distance R/2 from the centre. The electric potential at these points are such that

  • VA = VB = VC (b) VA = VB ¹ VC                    (c) VA ¹ VB ¹ VC  (d) VA ¹ VB = VC

Solution: (a)     Potential inside a conductor is always constant and equal to the potential at the surface.

 

Example: 36      Equal charges of

10 ´ 10 -9

3

coulomb are lying on the corners of a square of side 8 cm. The electric

 

potential at the point of intersection of the diagonals will be

 

(a) 900 V                                     (b)

Solution: (d)     Potential at the centre O

900      V

(c)

150     V

(d)

1500     V

 

 

V = 4 ´

1    .    Q

given Q =

Q                                                      Q

10 ´ 10 9 C  Þ  a  = 8 c+m =  8– ´ 10 2 m

 

4pe 0                               3

10 ´ 10 -9

+                  –

+                        –

+            C                                           + Q

 

V = 5 ´ 9 ´ 109 ´ 3             = 1500

                

8 ´ 10 -2

2 volt

+                         –

+       R             

+          –         r                                    Q                                                       Q

 

 

 

 
Tricky example: 4
A point charge q is placed at a distance of r from the centre of an uncharged conducting sphere of radius R (< r). The potential at any point on the sphere is

(a) Zero                           (b)    . q                                 (c)    . qR                     (d)   1 . qr 2

4pe 0 r                                           4pe 0 r 2                                 4pe 0      R

Solution: (c)     Since, potential V is same for all points of the sphere. Therefore, we can calculate its value at the centre of the sphere.

\ V =      1    . q + V ‘ ; where V¢ = potential at centre due to induced charge = 0 (because net

4pe 0 r

induced charge will be zero)  \ V =            1    . q .

4pe 0 r

Potential Due to Concentric Spheres.

To find potential at a point due to concentric sphere following guideline are to be considered

Guideline 1: Identity the point (P) at which potential is to be determined.

Guideline 2: Start from inner most sphere, you should know where point (P) lies w.r.t. concerning sphere/shell (i.e. outside, at surface or inside)

Guideline 3: Then find the potential at the point (P) due to inner most sphere and then due to next and so

on.

Guideline 4: Using the principle of superposition find net potential at required shell/sphere.

 

 

 

 

Case (ii) : The figure shows three conducting concentric shell of radii a, b and c (a < b < c) having charges Qa, Qb and Qc respectively what will be the potential of each shell

After following the guidelines discussed above                                                                                    Qc

Potential at A; V    =     é Qa   + Qb   + Qc ù                                                                                       Q  Qb

A       4pe   ê a        b         c ú                                                                                      a

0 ë                             û                                                                         c

   1   é Q         Q        Q  ù                                                                                b         a   A    B

Potential at B; VB =              ê   a +    b +    c ú

4pe 0 ë b           b         c û

Potential at C;V       1    é Qa  Qb  Qc ù C                    4pe 0 ê c       c                        c ú

ë                             û

 
Case (iii) : The figure shows two concentric spheres having radii r1 and r2 respectively (r2 > r1). If charge on inner sphere is +Q and outer sphere is earthed then determine.

(a)  The charge on the outer sphere

(b)   Potential of the inner sphere                                                                                                            Q¢      +Q

1     Q          1     Q’                                                                   r2

(i)   Potential at the surface of outer sphere V2 = 4pe . r + 4pe . r  = 0                                                r

0     2             0     2                                                               1

Þ Q’ = –Q

(ii)   Potential of the inner sphere V =  1 . Q +  1          (Q) =      Q     é 1 – 1 ù

1      4pe 0 r1       4pe 0 r2           4pe 0 ê r1        r2 ú

ë              û

Case (iv) : In the case III if outer sphere is given a charge +Q and inner sphere is earthed then

(a)   What will be the charge on the inner sphere

(b)  What will be the potential of the outer sphere

(i)      In this case potential at the surface of inner sphere is zero, so if Q’ is the charge induced on inner sphere

then V   =      é Q’ + Q ù = 0 i.e.,       Q’ = – r1 Q

1     4pe 0 ê r1      r2 ú                                  r

ë              û                                   2                                                                                             r2             +Q

(Charge on inner sphere is less than that of the outer sphere.)

(ii)   Potential at the surface of outer sphere                                                                                       r1

V2 =     . Q’    . Q

4per2        4per2

V        1      é- Q r1 + Qù  =     Q          é – r1 ù

2      4pe r   ê        r           ú       4pe r   ê1   r ú

0 2 ë          2         û               0 2 ë        2 û

Example: 37      A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 volts. The potential at the centre of the sphere is

  • Zero (b) 10 V

(c) Same as at a point 5 cm away from the surface                  (d) Same as at a point 25 cm away from the surface

Solution: (b)     Inside the conductors potential remains same and it is equal to the potential of surface, so here potential at the centre of sphere will be 10 V

Example: 38      A sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner sphere is charged to a potential 3 e.s.u. When the outer sphere is earthed. The charge on the inner sphere is

 

 

 

 

(a) 54 e.s.u.                                (b)

1 e.s.u.                             (c)  30 e.s.u.                         (d) 36 e.s.u.

4

 

Solution: (d)     Let charge on inner sphere be +Q. charge induced on the inner surface of outer sphere will be –Q. So potential at the surface of inner sphere (in CGS)

3 = Q Q

4     6

Þ Q = 36 e.s.u.

 

 

Example: 39      A charge Q is distributed over two concentric hollow spheres of radii r and (R > r) such that the surface densities are equal. The potential at the common centre is                                                                                 [IIT-JEE 1981]

 

 

(a)

Q(R2 + r 2)

 

4pe0(R + r)

(b)

   Q R + r

  • Zero (d)

     Q(R + r)    4pe0(R2 + r 2)

 

Solution: (d)     If q1 and q2 are the charges on spheres of radius r and R respectively, in accordance with conservation of charge

 

Q = q1

+ q2

….(i)                               q2

 

and according to the given problem s1 = s 2

 

q1           q2

 

q1       r 2

 

i.e.,

4pr 2 = 4pR2 Þ

q   R2

…. (ii)

 

2

So equation (i) and (ii) gives

q1 =

Qr2

 

(R2 + r 2)

and q2

=     QR2

(R2 + r 2)

 

 

Potential at common centre

V     é q1 + q2 ù =     é        Qr       +     QR        ù =     . Q(R + r)

 

ë
û

4pe0 êë r         R úû     4pe0 ê(R2 + r 2)       (R2 + r 2)ú        4pe0 (R2 + r 2)

Example: 40 A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 3Q, the new potential difference between the two surfaces is

(a) V                                             (b) 2V                                       (c) 4V                                   (d) –2V

Solution: (a) If a and b are radii of spheres and spherical shell respectively, potential at their surfaces will be

Vsphere =     . Q    and    Vshell =     . Q

4pe0 a                               4pe0 b

and so according to the given problem.

 

V Vsphere – Vshell

= Q é1 4pe0 êë a

1

ù

b úû

…. (i)

 

Now when the shell is given a charge –3Q the potential at its surface and also inside will change by

V0 =     é- 3Q ù

4pe0 êë     b úû

 

 

So         that        now

Vsphere =     é Q – 3Q ù

and

Vshell =     é Q – 3Q ù

hence

 

 

Vsphere – Vshell =    Q    é1 – 1 ù = V

4pe0 êë a

b úû

4pe0 êë b

b úû

 

4pe0 êë a       b úû

 

Example: 41      Three concentric metallic spheres A, B and C have radii a, b and c

(a < b < c)

and surface charge

 

densities on them are s , – s

and s respectively. The valves of VA

and

VB will be

 

 

è
ø

s                 s æ a2           ö

 

(a)

(a b c),       ç      – b + c ÷

 

 

 

 

(b)

e0

(a b c), a2

c

e0

e0 ç b               ÷

 

s

c
s

e0 æ a2               ö

 

(c)

(abc),

ç      – b + c ÷

 

s

s æ a2

ç                   ÷

è                   ø

÷

b2      ö            s

 

e0 ç c

(d)

ç      –

è            c

  • c ÷

ø

and

(a b + c)

e

0

 

Solution: (a)      Suppose charges on A, B and C are qa , qb and qc

 

 

Respectively, so s A = s = qa 

4pa2

Þ qa = s ´ 4pa2 , s B = –s =  qb    Þ qb = –s ´ 4pb2

4pb2

 

 

and s C = s =

qc

4pc2

Þ qc = s ´ 4pc2

 

Potential at the surface of A

 

 

 

V = (V )

+ (V )

+ (V )

    é qa qb qc ù

=  1  és ´ 4pa2 + (-s )´ 4pb2s ´ 4pc2 ù

 

 

 

  • A surface B  in         C  in

 

 

VA = s [a b c]]

4pe0

êë a      b

c úû

4pe

ê

0 êë       a                     b

ú

c        úû

 

e0

Potential at the surface of B

 

 

 

V = (V )

+ (V )

+ (V )

=  1 é qa qb qc ù =  1 és ´ 4pa2 – s ´ 4pb2 + s ´ 4pc2 ù

 

 

 

  • A out
  • surface
  • in

4pe0

êë b      b

c úû

4pe

ê

0 êë                     b                 b

ú

c       úû

 

ë
û

= s é a2 – b + cù

e0 ê b               ú

 

Electric Lines of Force.

(1)    Definition : The electric field in a region is represented by continuous lines (also called lines of force). Field line is an imaginary line along which a positive test charge will move if left free.

Electric lines of force due to an isolated positive charge, isolated negative charge and due to a pair of charge are shown below

 

 

  • Properties of electric lines of force

(i)  Electric field lines come out of positive charge and go into the negative charge.

  • Tangent to the field line at any point gives the direction of the field at that

(iii)   Field lines never cross each other.

  • Field lines are always normal to conducting

(v)  Field lines do not exist inside a conductor.

  • The electric field lines never form closed (While magnetic lines of forces form closed loop)

(vii)   The number of lines originating or terminating on a charge is proportional to the magnitude of charge. In the following figure electric lines of force are originating from A and terminating at B hence QA is positive while QB is negative, also number of electric lines at force linked with QA are more than those linked with QB hence |QA |>|QB |

 

 

 

 

 

  • Number of lines of force per unit area normal to the area at a point represents magnitude of intensity (concept of electric flux e., f = EA )

(ix)   If the lines of forces are equidistant and parallel straight lines the field is uniform and if either lines of force are not equidistant or straight line or both the field will be non uniform, also the density of field lines is proportional to the strength of the electric field. For example see the following figures.

 

 

  • Electrostatic shielding : Electrostatic shielding/screening is the phenomenon of protecting a certain region of space from external electric Sensitive instruments and

appliances are affected seriously with strong external electrostatic fields. Their working suffers and they may start misbehaving under the effect of unwanted fields.

r

E

The electrostatic shielding can be achieved by protecting and enclosing the sensitive instruments inside a hollow conductor because inside hollow conductors, electric fields is zero.

(i)   It is for this reason that it is safer to sit in a car or a bus during lightening rather than to stand under a tree or on the open ground.

  • A high voltage generator is usually enclosed in such a cage which is This would prevent the electrostatic field of the generator from spreading out of the cage.

(iii)   An earthed conductor also acts as a screen against the electric field. When conductor is not earthed field of the charged body C due to electrostatic induction

continues beyond AB. If AB is earthed, induced positive charge neutralizes and the field in the region beyond AB disappears.

 

 

 

 

Equipotential Surface or Lines.

If every point of a surface is at same potential, then it is said to be an equipotential surface

or

for a given charge distribution, locus of all points having same potential is called “equipotential surface” regarding equipotential surface following points should keep in mind :

  • The density of the equipotential lines gives an idea about the magnitude of electric Higher the density larger the field strength.

(2)   The direction of electric field is perpendicular to the equipotential surfaces or lines.

  • The equipotential surfaces produced by a point charge or a spherically charge distribution are a family of concentric

 

 

 

 

 

 

 

lines.

  • For a uniform electric field, the equipotential surfaces are a family of plane perpendicular to the field

 

(5)   A metallic surface of any shape is an equipotential surface e.g. When a charge is given to a metallic

 

surface, it distributes itself in a manner such that its every point comes at same potential even if the object is of irregular shape and has sharp points on it.

 

If it is not so, that is say if the sharp points are at higher potential then due to potential difference between these points connected through metallic portion, charge will flow from points of higher potential to points of lower potential until the potential of all points become same.

  • Equipotential surfaces can never cross each other

(7)   Equipotential surface for pair of charges

 

 

 

 

Example: 42      Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in

 

 

(a)

(b)

(c)

(d)

 

 

 

 

 

 

 

Solution (c) Option (a) shows lines of force starting from one positive charge and terminating at another. Option (b) has one line of force making closed loop. Option (d) shows all lines making closed loops. All these are not correct. Hence option (c) is correct

 

 

 

 

 

Example: 43 A metallic sphere is placed in a uniform electric field. The lines of force follow the path (s) shown in the figure as

 

 

 

 

 

(a) 1                                  (b) 2                                (c) 3                            (d) 4

Solution: (d) The field is zero inside a conductor and hence lines of force cannot exist inside it. Also, due to induced charges on its surface the field is distorted close to its surface and a line of force must deviate near the surface outside the sphere.

Example: 44  The figure shows some of the electric field lines corresponding to an electric field. The figure suggests

[MP PMT 1999]

 

 

 

 

 

 

Solution: (c)

(a)

EA  > EB  > EC

(b)

EA  = EB  = EC

(c)

EA  = EC  > EB

(d)

EA = EC  < EB

 

Example: 45 The lines of force of the electric field due to two charges q and Q are sketched in the figure. State if

  • Q is positive and Q > q

 

 

  • Q is negative and

 

  • q is positive and

Q  >  q

 

Q <  q

 

 

  • q is negative and Q < q

 

 

Solution: (c)     q is +ve because lines of force emerge from it and lines terminate at Q.

Q  <  q

because more lines emerge from q and less

 

Example: 46      The figure shows the lines of constant potential in a region in which an electric field is present. The magnitude of electric field is maximum at

  • A (b) B                                          (c)  C                               (d) Equal at A, B and C

Solution: (b)     Since lines of force are denser at B hence electric field is maximum at B

Example: 47      Some equipotential surface are shown in the figure. The magnitude and direction of the electric field is

(a) 100 V/m making angle 120o with the x-axis                       (b) 100 V/m making angle 60o with the x-axis

(c) 200 V/m making angle 120o with the x-axis                       (d) None of the above

 

Solution: (c)     By using dV = E dr cosq

suppose we consider line 1 and line 2 then

 

(30 – 20) = E cos 60o (20 – 10) × 10–2             So E = 200 volt / m making in angle 120o with x-axis

y

 

 

 

 

Relation Between Electric Field and Potential.

In an electric field rate of change of potential with distance is known as potential gradient. It is a vector quantity and it’s direction is opposite to that of electric field. Potential

gradient   relates   with   electric   field   according   to   the   following

 

relation E = – dV ;

dr

This relation gives another unit of electric field is

  volt .

meter

 

In the above relation negative sign indicates that in the direction of electric field potential decreases.

In   space  around   a   charge   distribution   we   can   also   write

r        ˆ      ˆ      ˆ

E = Ex i + Ey j + Ez k

 

 

where

E    = – dV ,

x           dx

Ey  = – dV

dy

and

E   = – dV

z           dz

 

 

 

With the help of formula

E = – dV , potential difference between any two points in an electric field can be

dr

 

determined by knowing the boundary conditions dV  = -òr2 E . dr  = -òr2 E . dr cosq .

r1                                 r1

For example: Suppose A, B and C are three points in an uniform electric field as shown in figure.

  • Potential difference between point A and B is

B

VB  VA  = -òA   E × dr

B
B

Since displacement is in the direction of electric field, hence q = 0o

 

 

So,

VB   VA   = -òA

E × dr cos 0 = – òA

E × dr = –Ed

 

 

In general we can say that in an uniform electric field

E = – V

d

or | E |= V

d

 

 

 

 

Another example

E = V

d

 

 

 

 

 

(ii)

C

Potential difference between points A and C is :

 

 

VC VA

= -òA  E dr cosq

= –E(AC) cosq

= –E(AB) = – Ed

 

Above relation proves that potential difference between A and B is equal to the potential difference between A and C i.e. points B and C are at same potential.

 

+      –

 

 

Example: 48      The electric field, at a distance of 20 cm from the centre of a dielectric sphere of radius 10 cm is 100 V/m.

The ‘E’ at 3 cm distance from the centre of sphere is                                                                          [RPMT 2001]

(a) 100 V/m                                (b) 125 V/m                             (c) 120 V/m                        (d) Zero

 

Solution: (c)     For dielectric sphere i.e. for non-conducting sphere Eout = k.q

r 2

and

Ein kqr

R 3

 

 

 

 

Eout

= 100          KQ         

(20 ´ 10-2 )2

Þ KQ = 100 ´ (0.2)2 so

Ein

= 100 ´ (0.2)2 ´ (3 ´ 10-2 )2 = 120 V/m

(10 ´ 10-2 )3

 

 

Example: 49      In xy co-ordinate system if potential at a point P(x, y) is given by V = axy ; where a is a constant, if r is

the distance of point P from origin then electric field at P is proportional to                                    [RPMT 2000]

(a) r                                              (b) r–1                              (c) r—2                          (d) r2

 

 

Solution: (a)     By using

E = – dV

dr

Ex   = – dV = –ay ,

dx

Ey = – dV = –ax dy

 

 

 

Electric field at point P E =

= a                    = ar

i.e., E µ r

 

 

Example: 50      The electric potential V at any point x, y, z (all in metres) in space is given by V = 4x2 volt. The electric

field at the point (1m, 0, 2m) in volt/metre is                                             [MP PMT 2001; IIT-JEE 1992; RPET 1999]

  • 8 along negative X-axis (b) 8 along positive X-axis
  • 16 along negative X-axis (d) 16 along positive Z-axis

 

 

Solution: (a)     By using

x-axis.

E = – dV

dx

Þ E = – d (4 x 2 ) = -8 x . Hence at point (1m, 0, 2m). E = – 8 volt/m i.e. 8 along – ve

dx

 

Example: 51                                                                                                                                                                                            The electric potential V is given as a function of distance x (metre) by V = (5x2 + 10x – 9) volt. Value of electric field at x = 1m is                                                                                                                                  [MP PET 1999]

(a) – 20 V/m                                   (b) 6 V/m                                 (c) 11 V/m                           (d) – 23 V/m

 

 

Solution: (a)     By using

E = – dV ;

dx

E = – d (5x 2 + 10x – 9) = (10x + 10) ,

dx

 

at             x = 1m            E = -20 V/m

 

Example: 52      A uniform electric field having a magnitude E0 and direction along the positive X-axis exists. If the electric potential V, is zero at X = 0, then, its value at X = +x will be                                                                         [MP PMT 1987]

(a) V(x)= +xE0                    (b) V(x)= – xE0                 (c) V(x)= x2E0                (d) V(x)= – x2E0

 

 

Solution: (b)     By using

E = – DV = – (V2  V1) ;      E   = – {V(x) 0}

Þ V(x) = – xE

 

Dr           (r2 – r1 )          0

x – 0                            0

 

Example: 53      If the potential function is given by V = 4x + 3y, then the magnitude of electric field intensity at the point (2, 1) will be                                                                                                                      [MP PMT 1999]

(a) 11                                 (b) 5                                (c) 7                            (d) 1

 

 

Solution: (b)     By using i.e., E =                       ; E

= – dV = – d (4 x + 3y) = -4

 

 

 

x          dx          dx

 

 

and

Ey = – dV = – d (4 x + 3y) = -3

 

dy          dy

 

 

\                       E =

= 5 N/C

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Work Done in Displacing a Charge.

(1)   Definition : If a charge Q displaced from one point to another point in electric field then work done

 

in this process is

W = Q ´ DV

where DV = Potential difference between the two position of charge Q.

 

 

( DV = E .D r

charge).

= EDr cosq

where q is the angle between direction of electric field and direction of motion of

 

 

 

  • Work done in terms of rectangular component of  E and   r   :      If charge Q is given a

 

displacement

r  = (r ˆi  + r  ˆj + r  kˆ)     in   an   electric   field     E  = (E ˆi  + E  ˆj + E

kˆ).

The   work  done   is

 

1        2         3                                                                              1           2           3

W = Q( E . r ) = Q(E1r1 + E2 r2 + E3 r3 ) .

 

 

Conservation of Electric Field.

As electric field is conservation, work done and hence potential difference between two point is path independent and depends only on the position of points between. Which the charge is moved.

 

 

 

 

Example: 54 A charge (– q) and another charge (+Q) are kept at two points A and B respectively. Keeping the charge (+Q) fixed at B, the charge (– q) at A is moved to another point C such that ABC forms an equilateral triangle of side l. The network done in moving the charge (– q) is

 

(a)

1    Qq

 

4pe 0 l

(b)

1    Qq

 

4pe 0 l 2

(c)

1

4pe 0

Qql

  • Zero

 

 

Solution: (d)     Since VA

= VC

kQ l

A  q

 

so   W = q (VC VA ) = 0

l                               l

 

 

B                                                            C

+ Q                    l

 

 

Example: 55      The work done in bringing a 20 coulomb charge from point A to point B for distance 0.2 m is 2 Joule. The potential difference between the two points will be (in volt)                                  [RPET 1999 Similar to MP PET 1999]

(a) 0.2                                (b) 8                                (c) 0.1                          (d) 0.4

 

Solution: (c)

W = Q.DV

Þ 2 = 20 ´ DV    Þ DV = 0.1 volt

 

Example: 56      A charge +q is revolving around a stationary +Q in a circle of radius r. If the force between charges is F

then the work done of this motion will be

[CPMT 1975, 90, 91, 97; NCERT 1980, 83; EAMCET 1994; MP PET 1993, 95;

MNR 1998; AIIMS 1997; DCE 1995; RPET 1998]

 

 

  • F × r (b)

F ´ 2pr

(c)

F

 

2pr

  • 0

 

Solution: (d)     Since +q charge is moving on an equipotential surface so work done is zero.

+ q

 

 

 

Example: 57      Four equal charge Q are placed at the four corners of a body of side ‘a’ each. Work done in removing a charge – Q from its centre to infinity is                                                                                      [AIIMS 1995]

 

 

  • 0 (b)

4pe 0 a

(c)

pe 0 a

(d)

Q2

 

2pe 0 a

 

Solution: (c)     We know that work done in moving a charge is W = QDV

Q                       a                  Q

 

Here

W = Q(V

V ) Q

A                                                        B

V   = 0 \ W = Q × V

 

 

 

Also V

0        ¥                           ¥                                                         0

 

= 4 ´  1 . Q     =                  =

 

a                 O                             a

 

0          4pe 0 a /           4pe 0a         pe 0a

Q

 

D                              a                         C Q                          Q

 

 

 

 

So, W =

pe 0 a

 

Example: 58      Two point charge 100 mC and 5 mC are placed at point A and B respectively with AB = 40 cm. The work done by external force in displacing the charge 5 mC from B to C, where BC = 30 cm, angle

 

ABC = p and    1    = 9 ´ 109 Nm2 /C 2

[MP PMT 1997]

 

2        4pe 0

 

 

  • 9 J (b)

81 J

20

(c)

  9 J

25

(d)

  • 9 J

4

 

Solution: (d)     Potential       at       B          due       to       +100        mC          charge       is         A

 

V    = 9 ´ 109 ´ 100 ´ 10-6

B                         40 ´ 10-2

= 9 ´ 106 volt

4

+ 100 mC

 

Potential       at       C          due       to       +100        mC          charge       is

 

VC = 9 ´ 109

´ 100 ´ 10-6 50 ´ 10-2

= 9 ´ 106

5

volt

50 cm

 

Hence    work   done    in    moving    charge    +5mC      from    B     to    C

W = 5 ´ 106 (VC VB )

p / 2

B

30 cm

+ 50 mC C

 

W = 5 ´ 10-6 æ 9 ´ 1069 ´ 10+6 ö = – 9 J

ç                                     ÷

5
4
4

è                                     ø

Example: 59      There is an electric field E in x-direction. If the work done in moving a charge 0.2 C through a

distance of 2 metres along a line making an angle 60o with the x-axis is 4J, what is the value of E [CBSE 1995]

 

  • 4 N/C (b) 8 N/C                                  (c)

N/C

  • 20 N/C

 

Solution: (d)     By using

W = q ´ DV

and DV = EDr cosq

 

So,

W = qE Dr cosq

W = 4 j = 0.2 ´ E ´ 2 ´ cos 60

 

Þ  E = 20 N/C                                                                                                                                                                      x

Example: 60      An electric charge of 20 mC is situated at the origin of XY co-ordinate system. The potential difference between the points. (5a, 0) and (– 3a, 4a) will be

  • a (b) 2a                                        (c) Zero                        (d)  a  

 

 

Solution: (c)

VA  kQ

5a

and VB  = kQ

5a

B (–3a, 4a)

 

\     VA

  • VB = 0

A (5a, 0)

 

 

 

Example: 61 Two identical thin rings each of radius R, are coaxially placed a distance R apart. If Q1 and Q2 are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is

æ Q1 ö

 

q(Q

Q )(

– 1)

q(Q

  • Q )

qç        ÷(

Q

– 1)

 

(a) Zero                              (b)

       1         2                

4pe 0 R

(c)

        1        2       

4pe 0 R

(d)

  è     2 ø              

4pe 0 R

 

 

 

Solution: (b)     Potential at the centre of first ring VA =

Q1      +

4pe0R

 

Q1                              Q2

 

 

Potential at the centre of second ring V

   Q2    +                Q1            

 

B

 

 

 

Potential difference between the two centres V

4pe 0 R

 

  • V

4pe 0                              1                                                   2

 

=

 

A             B

R

 

\ Work done W =

 

Equilibrium of Charge.

(1)   Definition : A charge is said to be in equilibrium, if net force acting on it is zero. A system of charges is said to be in equilibrium if each charge is separately in equilibrium.

  • Type of equilibrium : Equilibrium can be divided in following type:

(i)   Stable equilibrium : After displacing a charged particle from it’s equilibrium position, if it returns back then it is said to be in stable equilibrium. If U is the potential energy then in case of stable equilibrium

 

d 2U dx 2

is positive i.e., U is minimum.

 

  • Unstable equilibrium : After displacing a charged particle from it’s equilibrium position, if it never

 

returns back then it is said to be in unstable equilibrium and in unstable equilibrium maximum.

d 2U dx 2

is negative i.e., U is

 

 

(iii)   Neutral equilibrium : After displacing a charged particle from it’s equilibrium position if it neither comes back, nor moves away but remains in the position in which it was kept it is said to be in neutral

 

equilibrium and in neutral equilibrium

d 2U dx 2

is zero i.e., U is constant

 

  • Guidelines to check the equilibrium

(i)  Identify the charge for which equilibrium is to be analysed.

  • Check, how many forces acting on that particular

(iii)   There should be atleast two forces acts oppositely on that charge.

  • If magnitude of these forces are equal then charge is said to be in equilibrium then identify the nature of
  • If all the charges of system are in equilibrium then system is said to be in equilibrium

(4)  Different cases of equilibrium of charge

 

Case – 1 : Suppose three similar charge

Q1 , q

and

Case – 2 : Two similar charge

Q1   and  Q2

are

 

Q2 are placed along a straight line as shown below

placed along a straight line at a distance x from each other and a third dissimilar charge q is placed in between them as shown below

Charge q will be in equilibrium if | F1 | = | F2 |

 

Charge q will be in equilibrium if | F1 | = | F2 |

ç
÷

2

 

Q      æ x ö 2

i.e.,

 Q1    = æ x1 ö    .

 

i.e.,

      1    = ç   1 ÷

; This is the condition of

Q2            è x 2 ø

 

Q2            è x 2 ø

 

equilibrium of charge q. After following the guidelines we can say that charge q is in stable equilibrium and this system is not in equilibrium

Note : q Same short trick can be used here to find the position of charge q as we discussed in Case–1 i.e.,

 

Note : q

x =        x      

x =          x

and x =            x

 

1      1 + Q /Q

2         1

1      1 +   Q /Q

2         1

2      1 +   Q /Q

1        2

 

 

and

x 2 = 1 +

x

Q1 /Q2

□ It is very important to know that magnitude of charge q can be determined if one of the extreme

charge (either Q1 or Q2 ) is in equilibrium i.e. if Q2

 

2

e.g.   if two charges +4mC   and +16 mC   are

separated by a distance of 30 cm from each other

is in equilibrium then | q |= Q æ x 2 ö    and if Q   is

 

 

then for equilibrium a third charge should be placed

1 ç x ÷                  1

 

è
ø

between        them        at        a        distance

æ x   ö 2

 

in equilibrium then

| q |= Q2 ç 1 ÷

(It should be

 

x   =        30

= 10 cm or x

= 20 cm

è x ø

 

1    1 +

16 / 4                   2

remember that sign of q is opposite to that of Q1 (or Q2 ) )

 

 

 

(5)  Equilibrium of suspended charge in an electric field

(i)

Freely suspended charged particle : To suspend a charged a particle freely in air under the influence of electric field it’s downward weight should be balanced by upward electric force for example if a positive charge is suspended freely in an electric field as shown then

 

 

In equilibrium QE = mg

Þ E = mg

Q

 

Note : @In the above case if direction of electric field is suddenly reversed in any figure then acceleration of charge particle at that instant will be a = 2g.

  • Charged particle suspended by a massless insulated string (like simple pendulum) : Consider a charged particle (like Bob) of mass m, having charge Q is suspended in an electric field as shown under the influence of electric It turned through an angle (say q) and comes in equilibrium.

So, in the position of equilibrium (O¢ position)

 

T sinq

T cosq

= QE

= mg

….(i)

….(ii)

 

 

By squaring and adding equation (i) and (ii) T =

 

 

Dividing equation (i) by (ii)

tanq

QE mg

Þ   θ = tan -1 QE

mg

 

 

  • Equilibrium of suspended point charge system : Suppose two small balls having charge +Q on each are suspended by two strings of equal length l. Then for equilibrium position as shown in

 

 

 

 

 

T sinq = Fe

….(i)

 

T cosq

= mg

….(ii)

 

T 2 = (Fe  )2  + (mg)2

 

 

 

and

tanq = Fe

mg

; here

Fe  =

1   Q2

4pe 0 x 2

and

x = l sinq 2

 

 

  • Equilibrium of suspended point charge system in a liquid : In the previous discussion if point charge system is taken into a liquid of density r such that q remain same then

 

In equilibrium

Fe’ = T ‘ sinq

 

Fe’

and (mg Vrg) = T ‘ cosq

Q 2

 

\ tanq

=                    =

(mg – Vrg)   4pe 0

K (mgVrg)x 2

 

Fe                Q 2

 

When this system was in air tanq

=        =

mg       4pe 0 mgx 2

 

 

 

 

\ So equating these two gives us

1 =        1

m      k(m Vr)

Þ K =

m m Vr

=        1

æ1 – V r ö

 

m

ç             ÷

è             ø

If s is the density of material of ball then K =         1   

æ1 – θ ö

θ

ç        ÷

è         ø

QE Q

mg

 

Example: 62      A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium. If q is equal to

[CPMT 1999; MP PET 1999, MP PMT 1999; CBSE 1995; Bihar MEE 1995; IIT 1987]

 

  • Q

2

  • Q

4

 

æ x / 2 ö 2

  • + Q

4

  • + Q

2

 

Solution: (b)     By using Tricky formula

q Q ç           ÷

è   x   ø

 

Þ  q = Q

4

 

since q should be negative so q = – Q .

4

Tags:

Leave a Reply

Your email address will not be published. Required fields are marked *