Chapter 3 Electric Charges and Fields (Electrostatics Part 3) – Physics free study material by TEACHING CARE online tuition and coaching classes
Example: 63 Two point charges +4q and +q are placed at a distance L apart. A third charge Q is so placed that all the three charges are in equilibrium. Then location and magnitude of third charge will be
 At a distance Lfrom +4q charge, 4q
3 9
 At a distance
L from +4q charge, – 4q
3 9
 At a distance
2L from +4q charge, – 4q
 At a distance
3
2L from +q charge,
3
9
 4q 9
Solution: (c) Let third charge be placed at a distance
x_{1} from +4q charge as shown
Now x
= L = 2L
Þ x = L Q +q
1 3 2 3
1 +
+4q
x_{1} x_{2}
L
æ L / 3 ö ^{2} 4q 4q
For equilibrium of q, Q = +4qç ÷
è L ø
= 9 Þ Q = – 9 .
Example: 64 A drop of
10^{6} kg water carries
10^{6} C charge. What electric field should be applied to balance it’s
weight (assume g = 10 m/sec^{2})
(a)
10 V / m, Upward (b)
10 V / m, Downward (c) 0.1 V/m Downward (d)
0.1V / m,
Upward
Solution: (a) In equilibrium QE = mg
E = mg
Q
= 10^{6} ´ 10 10^{6}
= 10 V/m; Since charge is positive so electric field will be upward.
Example: 65 A charged water drop of radii 0.1 mm is under equilibrium in some electric field. The charge on the drop
is equivalent to electronic charge. The intensity of electric field is [RPET 1997]
(a)
1.61 N / C
(b)
25.2 N / C
æ 4 pr ^{3} r ö.g
(c)
262 N / C
(d)
1610 N / C
mg ç 3 ÷ 4 ´ (3.14)(0.1 ´ 10 ^{–}^{6} )^{3} ´ 10^{3} ´ 10
Solution: (c) In equilibrium QE = mg ; E =
= è ø = = 262 N/C
Q Q 3
1.6 ´ 10 ^{–}^{19}
Example: 66 The bob of a pendulum of mass 8 mg
carries an electric charge of
39.2 ´ 10 ^{10}
coulomb in an electric
Solution: (b)
field of 20 ´ 10^{3} volt / meter and it is at rest. The angle made by the pendulum with the vertical will be (a) 27^{o} (b) 45^{o} (c) 87^{o} (d) 127^{o}
T sinq = qE , T cosq = mg
qE
\ tanq =
mg
qE
mg
tanq = 39.2 ´ 10^{10} ´ 20 ´ 10^{3} = 1
8 ´ 10^{6} ´ 9.8
Þ q = 45^{o}
Example: 67 Two small spherical balls each carrying a charge Q = 10 mC (10 microcoulomb) are suspended by two
insulating threads of equal lengths 1 m each, from a point fixed in the ceiling. It is found that in equilibrium threads are separated by an angle 60^{o} between them, as shown in the figure. What is the
1
tension in the threads. (Given : (
) = 9 ´ 10^{9} Nm / C ^{2} )
 18 N
 8 N
(c) 0.18 N
(d) None of these
Solution: (b) From the geometry of figure
r = 1m
4pe _{0}
In the condition of equilibrium T sin 30^{o} = Fe
0o
T ´ 1
2
= 9 ´ 10^{9}.
(10 ´ 10 ^{6} )^{2}
12
+10 mC
Þ T= 1.8 N
Example: 68 Two similar balloons filled with helium gas are tied to L m long strings. A body of mass m is tied to another ends of the strings. The balloons float on air at distance r. If the amount of charge on the balloons is same then the magnitude of charge on each balloon will be
é mgr ^{2}
ù1 / 2
(a)
ê 2k


é 2k
tanq ú
úû
ù1 / 2
(b) ê mgr 2 tanq ú
(c)
ë
é mgr
êë 2k
é 2k
û
ù1 / 2
cotq úû
ù1 / 2
(d) ê mgr tanq ú
ë û
Solution: (a) In equilibrium R R
r
2R = mg
…. (i)
Fe = T sinq …. (ii)
R = T cosq
…. (iii) F
T sin
From equation (i) and (iii)
T cos
T
L q q L
2T cosq = mg
…. (iv)
m
Dividing equation (ii) by equation (iv) m
Q2
 F 1 k 2
æ mgr ^{2}
ö1 / 2


tanq = ^{e} Þ tanq = r Þ q = ç tanq ÷
 mg 2 mg
ç 2k ÷
Time Period of Oscillation of a Charged Body.
(1) Simple pendulum based : If a simple pendulum having length l and mass of bob m oscillates about it’s mean position than it’s time period of oscillation T = 2π
Case – 1 : If some charge say +Q is given to bob and an electric field E is applied in the direction as shown in figure then equilibrium position of charged bob (point charge) changes from O to O¢.
E d q
l O¢ QE mg¢ O mg
On displacing the bob from it’s equilibrium position 0¢. It will oscillate under the effective acceleration g¢, where mg’ = (mg)2 + (QE)2
Þ g’ = g ^{2} + (QE / m)2
Hence the new time period is T1 = 2p l g’
T1 = 2π l (g 2 + (QE/m)2 )2 1
Since g’ >g, hence T_{1} < T i.e. time period of pendulum will decrease. 
Case – 2 : If electric field is applied in the downward direction then.
Effective acceleration g’ = g + QE / m q l T So new time period E T2 = 2π l g + (QE/m) mg + QE T_{2} < T 
Case – 3 : In case 2 if electric field is applied in upward
direction then, effective acceleration. g’ = g – QE / m So new time period q QE l T3 = 2π l E g – (QE/m) T_{3} > T mg 

Case – 4 : In the case 3,
QE if T3 = T i.e., 2p l O¢¢ 2 (g – QE / m) 1 l E = 2 2p g Þ QE = 3 mg mg 

i.e., effective vertical force (gravity + electric) on the bob
= mg – 3 mg = – 2 mg, hence the equilibrium position O¢¢ of the bob will be above the point of suspension and bob will oscillate under on effective acceleration 2g directed upward. Hence new time period T4 = 2π l , T_{4} < T 2g 
 Charged circular ring : A thin stationary ring of radius R has a positive charge +Q If a negative charge – q (mass m) is placed at a small distance x from the centre. Then motion of the particle will be simple harmonic motion.
Electric field at the location of – q charge
E = 1
4pe _{0}
.
(x ^{2}
Qx

3
+ R ^{2} 2
Since x<< R, So
x ^{2} neglected hence
E = 1
4pe _{0}
. Qx R 3
Force experienced by charge – q is F = –q
1
4pe _{0}
. Qx R 3
Þ F µ –x hence motion is simple harmonic
Having time period T = 2π
(3) Spring mass system : A block of mass m containing a negative charge – Q is placed on a frictionless horizontal table and is connected to a wall through an unstretched spring of spring
constant k as shown. If electric field E applied as shown in figure the block experiences an electric force, hence spring compress and block comes in new position. This is called the equilibrium position of block under the influence of electric field. If block compressed further or stretched, it execute oscillation having time
period T = 2π
Neutral Point.
. Maximum compression in the spring due to electric field = QE
k
A neutral point is a point where resultant electrical field is zero. It is obtained where two electrical field are equal and opposite. Thus neutral points can be obtained only at those points where the resultant field is subtractive. Thus it can be obtained.
Zero Potential Due to a System of Two Point Charge.
If both charges are like then resultant potential is not zero at any finite point because potentials due to like charges will have same sign and can therefore never add up to zero. Such a point can be therefore obtained only at infinity.
If the charges are unequal and unlike then all such points where resultant potential is zero lies on a closed curve, but we are interested only in those points where potential is zero along the line joining the two charges.
Two such points exist, one lies inside and one lies outside the charges on the line joining the charges. Both the above points lie nearer the smaller charge, as potential created by the charge larger in magnitude will become equal to the potential created by smaller charge at the desired point at larger distance from it.
N
Example: 69 Two similar charges of +Q as shown in figure are placed at points A and B. – q charge is placed at point C midway between A and B. – q charge will oscillate if
 It is moved towards A
 It is moved towards B
 It is moved along CD
 Distance between A and B is reduced
Solution: (c) When – q charge displaced along CD, a restoring force act on it which causes oscillation.
Example: 70 Two point charges (+Q) and (– 2Q) are fixed on the Xaxis at positions a and 2a from origin respectively. At what position on the axis, the resultant electric field is zero
 Only x = 2 a
 Only
x = – 2 a
 Both
x = ± 2 a
x = 3a
2
only
Solution: (b) Let the electric field is zero at a point P distance d from the charge +Q so at P.
k.Q + k(–2Q) = 0
Þ 1 = 2
Þ d = a
d ^{2} (a + d)^{2}
d ^{2} (a + d)^{2}
( – 1)
Since d > a i.e. point P must lies on negative xaxis as shown at a distance x from origin hence
x = d – a =
a – a =
2a . Actually P lies on negative xaxis so x = –
2 a .
Example: 71 Two charges 9e and 3e are placed at a distance r. The distance of the point where the electric field intensity will be zero is [MP PMT 1989]
(a)
(c)
from 9e charge (b)
from 3e charge (d)
r from 9e charge
r from 3e charge
Solution: (b) Suppose neutral point is obtained at a distance x_{1} from charge 9e and x _{2} from charge 3e
By using
x r r


x1 =
æ 1 ö
1 + 1 +
ç1 + ÷ r
è 3 ø
Example: 72 Two point charges – Q and 2Q are separated by a distance R, neutral point will be obtained at
 A distance of
 A distance of
 A distance of
R from – Q charge and lies between the charges.
R from – Q charge on the left side of it
R from 2Q charge on the right side of it
 A point on the line which passes perpendicularly through the centre of the line joining – Q and 2Q
Solution: (b) As already we discussed neutral point will be obtained on the side of charge which is smaller in magnitude i.e. it will obtained on the left side of – Q charge and at a distance.
l = R
– 1
Þ l = R
Example: 73 A charge of + 4mC is kept at a distance of 50 cm from a charge of – 6mC. Find the two points where the potential is zero
 Internal point lies at a distance of 20 cm from 4mC charge and external point lies at a distance of 100
cm from 4mC charge.
 Internal point lies at a distance of 30 cm from 4mC charge and external point lies at a distance of 100
cm from 4mC charge
 Potential is zero only at 20 cm from 4mC charge between the two charges
 Potential is zero only at 20 cm from – 6mC charge between the two charges
Solution: (a) For internal point X,
x1 = x = 50 = 20cm
and for external point Y,

æ Q2 Q
ö
+ 1÷
6 + 1
4
x = x = 50 = 100cm
è 1 ø
^{1} æ Q ö 6
ç^{ } ^{2} – 1÷
4 – 1
è Q1 ø
100cm
50cm
Tricky example: 9
Two equal negative charges – q are fixed at points (0, a) and (0, – a) on the yaxis. A positive charge Q
is released from rest at the point (2a, 0) on the xaxis. The charge Q will
[IITJEE 1984, Bihar MEE 1995, MP PMT 1996]
 Execute simple harmonic motion about the origin
 Move to the origin and remains at rest
 Move to infinity
 Execute oscillatory but not simple harmonic
Solution: (d) By symmetry of problem the components of force on Q due to charges at A and B along yaxis will cancel each other while along xaxis will add up and will be along CO. Under the action of this force charge Q will move towards O. If at any time charge Q is at a distance x from O.
F Þ 2F cosq = 2 1 –qQ ´ x
A – q
i.e., F = –
4pe 0 (a ^{2} + x ^{2} )
1 . 2qQx
(a ^{2} + x ^{2} )^{1} ^{2}
a
4pe _{0}
(a ^{2} + x ^{2} )3 2
O q Q
x C
As the restoring force F is not linear, motion will be oscillatory (with a 2a
amplitude 2a) but not simple harmonic.
– q
B
Electric Potential Energy.
(1) Potential energy of a charge : Work done in bringing the given charge from infinity to a point in the electric field is known as potential energy of the charge. Potential can also be written as potential energy
per unit charge. i.e.
V = W = U .
Q Q
 Potential energy of a system of two charges : Since work done in bringing charge Q_{2} from ¥ to
point B is W = Q V , where V
is potential of point B due to charge Q
i.e. V
= 1 Q_{1}
So,
2 B B
W = U2 =

1
4pe_{0}
. Q1Q2
r
1 B 4pe r
This is the potential energy of charge Q_{2}, similarly potential energy of charge Q_{1}
will be U_{1} =
1
4pe_{0}
. Q1Q2
r
Hence potential energy of Q_{1} = Potential energy of Q_{2} = potential energy of system
U = Q1Q2 )
r
U = k Q1Q2 (in C.G.S.
r
Note : @Electric potential energy is a scalar quantity so in the above formula take sign of Q_{1} and Q_{2}.
 Potential energy of a system of n charges : In a system of n charges electric potential energy is calculated for each pair and then all energies so obtained are added i.e.
U = 1 é Q1Q2 + Q2Q3 +……….. ù
and in case of continuous distribution of charge. As
dU = dQ.V Þ


4pe_{0} ê
U = ò V dQ
r12
r23 ú
e.g. Electric potential energy for a system of three charges


Potential energy = 1 é Q1Q2 + Q2Q3 + Q3Q1 ù
4pe_{0} ê
r12
r23
r31 ú
While potential energy of any of the charge say Q
is 1 é Q1Q2 + Q3Q1 ù


^{1} 4pe ê
r12
r31 ú
Note : @For the expression of total potential energy of a system of n charges consider pair of charges.
n(n – 1)
2
number of
 Electron volt (eV) : It is the smallest practical unit of energy used in atomic and nuclear As
electron volt is defined as “the energy acquired by a particle having one quantum of charge 1e when
accelerated by 1volt” i.e. 1eV = 1.6 ´10^{19} C ´ 1J
C
= 1.6 ´ 10^{19} J
= 1.6 ´ 10^{–12} erg
volt)
Energy acquired by a charged particle in eV when it is accelerated by V volt is E = (charge in quanta) × (p.d. in
Commonly asked examples :
S.No.  Charge  Accelerated by p.d.  Gain in K.E. 
(i)  Proton  5 ´ 10^{4} V  K = e ´ 5 ´ 10^{4} V = 5 ´ 10^{4} eV = 8 ´ 10^{–15} J [JIPMER 1999] 
(ii)  Electron  100 V  K = e ´ 100 V = 100 eV = 1.6 ´ 10^{–17} J [MP PMT 2000; AFMC 1999] 
(iii)  Proton  1 V  K = e ´ 1 V = 1 eV = 1.6 ´ 10^{–19} J [CBSE 1999] 
(iv)  0.5 C  2000 V  K = 0.5 ´ 2000 = 1000 J [JIPMER 2002] 
(v)  aparticle  10^{6} V  K = (2e) ´ 10^{6} V = 2 MeV [MP PET/PMT 1998] 
 Electric potential energy of a uniformly charged sphere : Consider a uniformly charged sphere of radius R having a total charge Q. The electric potential energy of this sphere is equal to the work done in bringing the charges from infinity to assemble the
U = 3Q^{2}
20pe_{0}R
(6) Electric potential energy of a uniformly charged thin spherical shell :
U = Q2
8pe_{0} R
(7) Energy density : The energy stored per unit volume around a point in an electric field is given by
Ue =
U
Volume
= 1 e
2 0
E^{2} . If in place of vacuum some medium is present then Ue
= 1 e
2
_{0}e _{r} E ^{2}
Q q
U = 0
Example: 74 If the distance of separation between two charges is increased, the electrical potential energy of the system [AMU 1998]
(a) May increases or decrease (b) Decreases
(c) Increase (d) Remain the same
Solution: (a) Since we know potential energy
U = k. Q1Q2
r
As r increases, U decreases in magnitude. However depending upon the fact whether both charges are similar or disimilar, U may increase or decrease.
Example: 75 Three particles, each having a charge of 10mC are placed at the corners of an equilateral triangle of side
10cm. The electrostatic potential energy of the system is (Given 1 = 9 ´ 10^{9} N – m^{2} /C ^{2} )
4pe _{0}
(a) Zero (b) Infinite (c) 27 J (d) 100 J
[AMU 1998]
Solution: (c) Potential energy of the system,
U = 9 ´ 10^{9} é(10 ´ 10 6 )2 ´ 3ù
= 27 J
10mC




ê 0.1 ú
10mC
10 cm
10 cm
10 cm
10mC
Example: 76 Three charges Q, +q and +q are placed at the vertices of a rightangled isosceles triangle as shown. The
net electrostatic energy of the configuration is zero if Q is equal to [IIT (Screening) 2000]
 –q
 – 2 q
(d) +q
Solution: (b) Potential energy of the configuration U = k. Qq + k.q 2 + k. Qq = 0 Þ Q =
a a a
Example: 77 A charge 10 e.s.u. is placed at a distance of 2 cm from a charge 40 e.s.u. and 4 cm from another charge of 20 e.s.u. The potential energy of the charge 10 e.s.u. is (in ergs)
(a) 87.5 (b) 112.5 (c) 150 (d) 250
Solution: (d) Potential energy of 10 e.s.u. charge is
U = 10 ´ 40 + 10 ´ 20 = 250 erg.
10 esu
2 4
20 esu
Example: 78 In figure are shown charges q_{1} = + 2 × 10^{–8} C and q_{2} = – 0.4 × 10^{–8} C. A charge q_{3} = 0.2 × 10^{–8} C in moved along the arc of a circle from C to D. The potential energy of q_{3} [CPMT 1986]
 Will increase approximately by 76%
 Will decreases approximately by 76%
 Will remain same
 Will increases approximately by 12%
Solution: (b) Initial potential energy of q U = æ q1q3 + q2 q3 ö ´ 9 ´ 10^{9} q

3 i ç
è
0.8 1 ÷ 3
æ q q q q ö C
Final potential energy of q_{3} U _{f} = ç ^{1} ^{3} + ^{2} ^{3} ÷ ´ 9 ´ 10^{9}
è
Change in potential energy = U_{f} – U_{i}
0.8
0.2 ø
Now percentage change in potential energy = U f – Ui ´ 100
ui
A B
D
q1 60 cm q2
80 cm
q q æ 1 – 1ö ´ 100


2 3 ç 0.2 ÷
=
On putting the values ~– – 76%

q æ q_{1}
è
+ q_{2} ö

ø
Motion of Charged Particle in an Electric Field.
(1) When charged particle initially at rest is placed in the uniform field :
Let a charge particle of mass m and charge Q be initially at rest in an electric field of strength E
Fig. (A) Fig. (B)
 Force and acceleration : The force experienced by the charged particle is F = QE . Positive charge
experiences force in the direction of electric field while negative charge experiences force in the direction opposite to the field. [Fig. (A)]
Acceleration produced by this force is a = F
m
= QE
m
Since the field E in constant the acceleration is constant, thus motion of the particle is uniformly accelerated.
 Velocity : Suppose at point A particle is at rest and in time t, it reaches the point B [Fig. (B)]
V = Potential difference between A and B; S = Separation between A and B
 By using v = u + at ,
v = 0 + Q E t , Þ
m
v = QEt
m
 By using v^{2} = u^{2} + 2as , v^{2} = 0 + 2 ´ QE´ s
m
v2 = 2QV
m
V





íQ v =
î
 Momentum : Momentum p = mv,
p = m ´ QEt = QEt m
or p = m ´ =
 Kinetic energy : Kinetic energy gained by the particle in time t is
K = 1 mv 2 = 1 m (QEt)^{2}
= Q 2 E 2 t 2 or
K = 1 m ´ 2QV
= QV
2 2 m 2m 2 m
(2) When a charged particle enters with an initial velocity at right angle to the uniform field :
When charged particle enters perpendicularly in an electric field, it describe a parabolic path as shown
 Equation of trajectory : Throughout the motion particle has uniform velocity along xaxis and horizontal displacement (x) is given by the equation x = ut
Since the motion of the particle is accelerated along y–axis, we will use equation of motion for uniform
acceleration to determine displacement y. From S = ut + 1 at ^{2}
2
We have u = 0
(along yaxis) so
y = 1 at ^{2}
2
i.e., displacement along yaxis will increase rapidly with time (since y µ t ^{2} )
x
From displacement along xaxis t = u
1 æ QE ö æ x ö ^{2}
So y = ç ÷ ç ÷ ; this is the equation of parabola which shows y µ x ^{2}
2 è m ø è u ø
(ii) Velocity at any instant : At any instant t, vx
= u and vy
= QEt
m
So v = v= =
If b is the angle made by v with xaxis than tan b = vy = QEt .
vx mu
r
Example: 79 An electron (mass =
9.1 ´ 10 ^{31} kg
and charge =
1.6 ´ 10 ^{–}^{19} coul.)
is sent in an electric field of
intensity 1 ´ 10^{6} V / m. How long would it take for the electron, starting from rest, to attain one–tenth the velocity of light
(a)
1.7 ´ 10 ^{–}^{12} sec
(b)
1.7 ´ 10 ^{–}^{6} sec
1.7 ´ 10 ^{–}^{8} sec
1.7 ´ 10 ^{–}^{10} sec
Solution: (b) By using v = QEt Þ 1 ´ 3 ´ 10^{8} = (1.6 ´ 10 19 ) ´ 106 ´ t Þ t = 1.7 ´ 10 ^{–}^{10} sec.
m 10
9.1 ´ 10 ^{31}
Example: 80 Two protons are placed
10 ^{–}^{10} m
apart. If they are repelled, what will be the kinetic energy of each
proton at very large distance
(a)
23 ´ 10 ^{19} J
(b)
11.5 ´ 10 ^{19} J
(c)
2.56 ´ 10 ^{19} J
(d)
2.56 ´ 10 ^{28} J
Solution: (d) Potential energy of the system when protons are separated by a distance of 10 ^{10} m is
9 19 2
U = 9 ´ 10 ´ (1.6 ´ 10 ) = 23 ´ 10 19 J
10 10
P+ p+
According to law of conservation of energy at very larger distance, this energy is equally distributed in both the protons as their kinetic energy hence K.E. of each proton will be 11.5 ´ 10 ^{19} J.
Example: 81 A particle A has a charge +q and particle B has charge +4q with each of them having the same mass m. When allowed to fall from rest through the same electrical potential difference, the ratio of their speeds
v A will becomes [BHU 1995; MNR 1991]
vB
(a) 2 : 1 (b) 1 : 2 (c) 1 : 4 (d) 4 : 1
Solution: (b) We know that kinetic energy
K = 1 mv^{2} = QV . Since, m and V are same so,
2
v^{2} µ Q Þ
vA =
vB
= = 1 .
2
Example: 82 How much kinetic energy will be gained by an a – particle in going from a point at 70 V to another point
at 50 V [RPET 1996]
 40 eV (b) 40 keV (c) 40 MeV (d) 0 eV
Solution: (a) Kinetic energy K = QDV
Þ K = (2e)(70 – 50) V
= 40 eV
Example: 83 A particle of mass 2g and charge 1m C is held at a distance of 1 metre from a fixed charge of 1mC. If the particle is released it will be repelled. The speed of the particle when it is at a distance of 10 metres from
the fixed charge is [CPMT 1989]
(a) 100 m/s (b) 90 m/s (c) 60 m/s (d) 45 m/s
Solution: (b) According to conservation of energy
1 mC 1 mC
A
Fixed charge 1m
Moving charge B
10m
Energy of moving charge at A = Energy of moving charge at B
9 ´ 10^{9} ´ 103 ´ 106 = 9 ´ 10^{9} ´ 103 ´ 106 + 1 ´(2 ´ 10^{–}^{3}) v^{2}
1 10 2
Þ v^{2} = 8100 Þ v = 90 m/sec
Force on a Charged Conductor.
To find force on a charged conductor (due to repulsion of like charges) imagine a small part XY to be cut and just separated from the rest of the conductor MLN. The field in the cavity due to the rest of the conductor is E_{2}, while field due to small part is E_{1}. Then
Inside the conductor E = E1 – E2 = 0 or
s
E1 = E2

Outside the conductor E = E1 + E2 =
0
s

Thus E1 = E2 = 2e
To find force, imagine charged part XY (having charge
s dA
placed in the cavity MN having field E_{2}).
Thus force dF = (s dA)E
or dF = s 2
dA . The force per unit area or electric pressure is
dF = s ^{2}

^{2} 2e dA 2e
0
The force is always outwards as (±s )^{2} will try to expand the charged body.
is positive i.e., whether charged positively or negatively, this force
A soap bubble or rubber balloon expands on given charge to it (charge of any kind + or –).
Equilibrium of Charged Soap Bubble.
For a charged soap bubble of radius R and surface tension T and charge density s . The pressure due to
surface tension 4 T
R
and atmospheric pressure
Pout act radially inwards and the electrical pressure
(Pel )
acts
radially outward.
The total pressure inside the soap bubble P = P
+ 4T – s ^{2}
in out
R 2e_{0}
Excess pressure inside the charged soap bubble P – P = P
= 4T – s ^{2}
. If air pressure inside and
in out
excess
R 2e_{0}
outside are assumed equal then
P = P
i.e., P
= 0 . So,
4T = s ^{2}
in out
excess
R 2e_{0}
This result give us the following formulae
8 e_{0} T
 Radius of bubble R = s 2
 Surface tension T = s2 R
8e_{0}
(3) Total charge on the bubble Q = 8pR
 Electric field intensity at the surface of the bubble E = =