Electrostatic Potential and Capacitance Difficult Questions and Answers | Page - 1

Electrostatic Potential and Capacitance

Three charges, each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC, 2a. D and E are the mid points of BC and CA. The work done in taking a charge Q from D to E is

1. 3qQ
  8πε₀a
1. qQ
  4πε₀a
1. zero
1. 3qQ
  4πε₀a

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AC = BC
V_D = V_E
We have,
W = Q (V_E – V_D)
⇒ W = 0

Correct Option: C

AC = BC
V_D = V_E
We have,
W = Q (V_E – V_D)
⇒ W = 0

As per the diagram, a point charge +q is placed at the origin O. Work done in taking another point charge – Q from the point A [coordinates (0, a)] to another point B [coordinates (a, 0)] along the straight path AB is:

1. zero
1. -qQ 1 √2a
  4πε₀ a²
1. qQ 1 . a
  4πε₀ a² √2
1. qQ 1 √2a
  4πε₀ a²

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We know that potential energy of two charge system is given by
U = 1 = q₁q₂
4πε₀ r

According to question,
U_A = 1 (+q)(-Q) = - 1 Qq
4πε₀ a 4πε₀ a

and U_B = 1 (+q)(-Q) = - 1 Qq
4πε₀ a 4πε₀ a

∆U = U_B – U_A = 0
We know that for conservative force,
W = –∆U = 0

Correct Option: A

We know that potential energy of two charge system is given by
U = 1 = q₁q₂
4πε₀ r

According to question,
U_A = 1 (+q)(-Q) = - 1 Qq
4πε₀ a 4πε₀ a

and U_B = 1 (+q)(-Q) = - 1 Qq
4πε₀ a 4πε₀ a

∆U = U_B – U_A = 0
We know that for conservative force,
W = –∆U = 0

Charges +q and –q are placed at points A and B respectively which are a distance 2L apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is

1. qQ
  2πε₀L
1. qQ
  6πε₀L
1. - qQ
  6πε₀L
1. qQ
  4πε₀L

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Potential at C = V_C = 0
Potential at D = V_D
= k -q + Kq = - 2 kq
L 3L 3 L

Potential difference
V_D - V_C = -2 kq = 1 - 2 . q
3 L 4πε₀ 3 L

⇒ Work done = Q (V_D - V_C)
= - 2 × 1 qQ = - qQ
3 4πε₀ L 6πε₀L

Correct Option: C

Potential at C = V_C = 0
Potential at D = V_D
= k -q + Kq = - 2 kq
L 3L 3 L

Potential difference
V_D - V_C = -2 kq = 1 - 2 . q
3 L 4πε₀ 3 L

⇒ Work done = Q (V_D - V_C)
= - 2 × 1 qQ = - qQ
3 4πε₀ L 6πε₀L

Two charges q₁ and q₂ are placed 30 cm apart, as shown in the figure. A third charge q3 is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is (q₃/4πε₀)k where k is

1. 8q₁
1. 6q₁
1. 8q₁
1. 6q₁

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We know that potential energy of discrete system of charges is given by
U = 1 q₁q₂ + q₂q₃ + q₃q₁
4πε₀ q₁₂ q₂₃ q₃₁

According to question,
U_initial = 1 q₁q₂ + q₂q₃ + q₃q₁
4πε₀ 0.3 0.5 0.4

U_final = 1 q₁q₂ + q₂q₃ + q₃q₁
4πε₀ 0.3 0.1 0.4

U_final - U_initial = 1 q₁q₂ - q₂q₃
4πε₀ 0.1 0.4

= 1 [10q₂q₃ - 2q₂q₃] = q₃ (8q₂)
4πε₀ 4πε₀

Correct Option: C

We know that potential energy of discrete system of charges is given by
U = 1 q₁q₂ + q₂q₃ + q₃q₁
4πε₀ q₁₂ q₂₃ q₃₁

According to question,
U_initial = 1 q₁q₂ + q₂q₃ + q₃q₁
4πε₀ 0.3 0.5 0.4

U_final = 1 q₁q₂ + q₂q₃ + q₃q₁
4πε₀ 0.3 0.1 0.4

U_final - U_initial = 1 q₁q₂ - q₂q₃
4πε₀ 0.1 0.4

= 1 [10q₂q₃ - 2q₂q₃] = q₃ (8q₂)
4πε₀ 4πε₀

The potential energy of particle in a force field is
U = A - B
r² r

, where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is :

1. B/2A
1. 2A/B
1. A/B
1. B/A

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for equilibrium
dU = 0 ⇒ -2A + B = 0
dr r³ r²

r = 2A
B

for stable equilibrium
d²U/dr² should be positive for the value of r.
here d²U = 6A - 2B is +ve value for r = 2A So
dr² r⁴ r³ B

Correct Option: B

for equilibrium
dU = 0 ⇒ -2A + B = 0
dr r³ r²

r = 2A
B

for stable equilibrium
d²U/dr² should be positive for the value of r.
here d²U = 6A - 2B is +ve value for r = 2A So
dr² r⁴ r³ B