Questions having one mark each: Questions having 2 marks each: Questions having 3 mark each: Questions having 5 marks each:
i. electric field intensity
ii. Electric dipole moment.
i. Find at what distance from the 1st charge, q1, would the electric potential be zero.
ii. Also calculate the electrostatic potential energy of the system.
i. parallel to the field,
ii. perpendicular to the field.
i. magnitude of electric field
ii. the potential energy of the dipole.
i. on the surface
ii. inside and
If two similar plates, each of area A having surface charge densities +σ and –σ are separated by a distance‘d’ in air, write the expressions for
i. The electric field at points between the two plates.
ii. The potential difference between the plates.
iii. The capacitance of the capacitor so formed.
i. the final potential difference across the combination.
ii. Initial and final energies. How will you account for the difference in energies?
iii. potential difference
iv. electric field
v. energy stored?
i. Show that no translatory force acts on it.
ii. Derive an expression for the torque acting on it.
iii. The dipole is aligned parallel to the field. Calculate the work done in rotating it through 1800 .
Questions having one mark each:
Questions having 2 marks each:
Questions having 3 mark each:
Questions having 5 marks each:
Questions: 1. An electrostatic charge of 21 u C is placed at a distance in air of 15 cm
From a second charge. The force of attraction between the two charges is 26 N.
A. Calculate the magnitude of the second charge.
Sol) q1 = 21 u C = 21 x 10-6 C
q2 = ?, F = 26N, R = 15cm = 0.15m
Using F = q1 q2 / 4πεoR2
1/4πεo = 9 x 109 N-m2-C-2
Thus F = 9 x 109 q1 q2 / ( 0.15)2
or 26 = 9 x 109 q1 q2 / ( 0.15)2
or 26 = 9 x 109 x 21 x 10-6 q2 / ( 0.15)2 = 1.89 x 105 q2 / ( 0.15)2
or q2 = 26 x ( 0.15)2 / 1.89 x 105
or q2 = 3.1 x 10-6 C = 3.1 μC
B If the distance between the charges is decreased to 5.0 cm, calculate the magnitude of the new force acting on each ball.
Sol) Force between the balls when distance is 15cm = F1 = 26N, R1 = 15cm
Let force between the balls when distance is 5cm be = F2 , R2 = 5cm
Using relation, F = q1 q2 / 4πεoR2
we have F1 = q1 q2 / 4πεoR1 2
and F2 = q1 q2 / 4πεoR2 2
Thus F1 / F2 = R2 2 / R1 2
or 26 / F2 = 52 / 152 = 25 / 225 = 1/9
or F2 = 26*9 = 234 N
2. An electron, moving through an electric field, experiences an acceleration of 6.3 x 103 m/s2.
A. How much electrostatic force is acting on the electron?
Sol) Acceleration a = 6.3 x 103 m/s2.
Charge of electron = e = 1.6 x 10-19 C
Mass of electron = 9.1 x 10-31 Kg
Let Electric force be = F
Now F = ma
or F = 9.1 x 10-31 x 6.3 x 103 = 5.73 x 10-27 N
B. What is the strength of the electric field?
Sol) Let strength of electric field be E
Now F = e E
but F = 5.73 x 10-27 N (as calculated above in part A)
or E = F/e = 5.73 x 10-27 / 1.6 x 10-19
or E = 3.58 x 10-8 N/C
4. A 3.75 x 10- 9 F capacitor carries a charge of 1.75x10 –8 C
A. What is the potential difference across the plates?
Sol) C = 3.75 x 10- 9 F
Q = 1.75x10 –8 C
Now Using Q = CV
we have V = Q/C = 1.75x10 –8 / 3.75 x 10- 9 = 4.67 V
B. If the plates are 6.50 x 10-4 m a part, what is the strength of the
Electric field between them?
Sol) Distance between the plates, d = 6.50 x 10-4 m
V = 4.67 V (as calculated above in part A)
Thus E = V/d = 4.67 / 6.50 x 10-4 = 7.18 x 103 V/m
5. A charge of –3.00 u C is fixed at the centre of a compass. Two
Additional charges are fixed on the circle of the compass (radius = 0.100m). The charges on the circle are –4.00 u C at the position due north and + 5.00 u C at a position due east. What is the magnitude and direction of the net electrostatic force acting on the charge at the centre?
Sol) q = –3.00 u C = -3 x 10-6 C
Here BC = AC = 0.1m
Here charge q is placed at C (shown in above figure)
Let at point B which is north to C charge put be = q1 = -4 uC = -4 x 10-6 C
also at point A which is east to C charge kept = q2 = 5 x 10-6 C
Forces at charge q placed at point C due to charges at B and A are mutually perpendicular to each other.
FB = 9 x 109 q.q1 / R2
or FB = 9 x 109 x (-3 x 10-6) x (-4 x 10-6) / (0.1)2
or FB = 10.8 N long BC
Similarly force on q due to q2 at B = FA = 9 x 109 q.q2 / R2
or FA = 9 x 109 x (-3 x 10-6) x (5 x 10-6) / (0.1)2
or FA = 13.5 N along CA
Thus total force F on charge q placed at C is given by
or F =
or F = 17.28 N
The direction of the force is such that it makes an angle θ with CA given by
θ = tan-1 FA/FB = tan-1 13.5/10.8 = tan-1 1.25
or θ = 51.34 degrees.
Hence the direction of the force F = 17.28 N is such that it makes an angle 51.34 deg with CA
6. A particle (mass = 2.5 x 10-15 kg) with a negative charge of 3.5 x 10-5 C is moved from the positive plate of a parallel plate capacitor to the negative plate, a distance of 2.5 cm.
A. If the potential difference between the two plates is 15 volts, how much work is done in moving the particle?
Sol) m = 2.5 x 10-15 kg
q = -3.5 x 10-5 C
Distance between the capacitor plates, d = 2.5 cm = 0.025 m
Potential difference between the plates, V = 15 volts
Thus intensity of electric field E = V/d = 15/0.025 = 600 V/m
Work done in moving the particle = W = F.d = qE.d
or W = qE.d = 3.5 x 10-5 x 600 x 0.025 = 5.25 x 10-4 J is the amount of work done in moving the particle.
B. If the particle is released from the negative plate, it will accelerate toward the positive plate. What will be the velocity of the particle just before it strikes the positive plate?
Sol) m = 2.5 x 10-15 kg
q = -3.5 x 10-5 C
V = 15 volts
E = 600V/m (as calculated above)
Now acceleration of the charged particle = a = qE/m
or a = 3.5 x 10-5 x 600 / 2.5 x 10-15
or a = 8.4 x 1012 m/s2
Using the relation v2 – u2 = 2aS
here S = d = 0.025 m
u = 0
Thus v2 – 0 = 2 x 8.4 x 1012 x 0.025
or v2 = 4.2 x 1011 m/s
or v = 6.48 x 105 m/s
C. What is the strength of the electric field between the plates?
Sol) Electric field between the plates E = V/d = 15/0.025
or E = 600 V/m
D. What force will be applied to the particle as it accelerates toward the Negative plate?
Sol) Force on the particle F is given by
F = qE
Here, q = -3.5 x 10-5 C
E = 600 V/m
Thus F = 3.5 x 10-5 x 600 N = 2.1 x 10-2 N = 0.021 N