Two point charges of + 1 . 0 × 10 - 6 C and - 1 . 0 × 10 - 6 C are at a distance 2 . 0 cm apart from each other. This electric dipole is situated in a uniform electric field of 1 . 0 × 10 5 V - m - 1 . What will be the maximum torque acting on it due to the field? What will be the work done for rotating it through 180 ∘ from the equilibrium position?

Given that q 1 = 1 . 0 × 10 - 6 C and q 1 = q 2 = q ⇒ q 2 = 1 . 0 × 10 - 6 C E = 1 × 10 5 NC - 1 2 r = 2 . 0 cm = 2 × 10 - 2 m r = 2 . 0 cm ∴ P = q 2 r = 1 × 10 - 6 × 2 × 10 - 2 C-m P = 2 × 10 - 8 C-m ∴ torque acting on q due to E → is 𝜏 = PEsinθ θ = 90 ° for max torque ⇒ 𝜏 = 2 × 10 - 8 × 1 × 10 + 5 = 2 × 10 - 3 N - m ∴ 𝜏 = 2 × 10 - 3 N - m Work done in rotating the dipole in uniform electric field . W = PE 1 - cosθ θ = 180 ° = 2 × 10 - 3 1 + 1 = 2 × 2 × 10 - 3 ∴ W = 4 × 10 - 3 J

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Two positive charges of $10 μC$ each are situated $10 cm$ apart from each other. What is the increase in the electrostatic potential energy of the system when the two charges are brought closer by $2 cm$ ?

Important Questions on Electric Potential

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NOOTAN ISC PHYSICS Class 12 Part-1>Chapter 3 - Electric Potential>QUESTIONS>Q 25

Calculate the work required to assemble each of the systems of charges (which is same as the electrostatic potential energy of the system) shown in Fig.

(a)

and

(b)

MEDIUM

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NOOTAN ISC PHYSICS Class 12 Part-1>Chapter 3 - Electric Potential>QUESTIONS>Q 25

(ii) A charge $q_{0}$ is brought from infinity to the centre of the square, the four charges being held fixed at its comers. How much extra work is needed to do this?

Question Image

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NOOTAN ISC PHYSICS Class 12 Part-1>Chapter 3 - Electric Potential>QUESTIONS>Q 26

The distance between two protons is

1.0 \times 10^{- 10} m

. If they are made free, what will be the kinetic energy of each when they are away at a very large distance from each other? If one proton is kept fixed and only other is free then what will be its kinetic energy?

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NOOTAN ISC PHYSICS Class 12 Part-1>Chapter 3 - Electric Potential>QUESTIONS>Q 27

Two electrons are released towards each other with equal velocities of

10^{6} m s^{- 1}

. What will be the closest approach between them?

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NOOTAN ISC PHYSICS Class 12 Part-1>Chapter 3 - Electric Potential>QUESTIONS>Q 28

Two point charges of

+ 1.0 \times 10^{- 6} C

and

- 1.0 \times 10^{- 6} C

are at a distance

2.0 cm

apart from each other. This electric dipole is situated in a uniform electric field of

1.0 \times 10^{5} V - m^{- 1}

. What will be the maximum torque acting on it due to the field? What will be the work done for rotating it through

180^{\circ}

from the equilibrium position?

HARD

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NOOTAN ISC PHYSICS Class 12 Part-1>Chapter 3 - Electric Potential>QUESTIONS>Q 29

An electric dipole of length

2 cm

is placed with its axis making an angle of

60^{\circ}

to a uniform electric field of

10^{5} N ‐ C^{- 1}

. If it experiences a torque of

8 \sqrt{3} N ‐ m,

calculate magnitude of the charge on the dipole.

EASY

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IMPORTANT

NOOTAN ISC PHYSICS Class 12 Part-1>Chapter 3 - Electric Potential>QUESTIONS>Q 29

An electric dipole of length

2 cm

is placed with its axis making an angle of

60 °

to a uniform electric field of

10^{5} N - C^{- 1}

. If it experiences a torque of

8 \sqrt{3} N - m

, calculate potential energy of the dipole.

HARD

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IMPORTANT

NOOTAN ISC PHYSICS Class 12 Part-1>Chapter 3 - Electric Potential>QUESTIONS>Q 30

Two point charges of

+ 3 μC

and

- 3 μC

are at a distance

3 \times 10^{- 3} m

apart from each other. Calculate electric potential at a distance of

0.6 m

from the dipole in broad-side-on position.

Two positive charges of 10 μC each are situated 10 cm apart from each other. What is the increase in the electrostatic potential energy of the system when the two charges are brought closer by 2 cm?

Important Questions on Electric Potential

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Two positive charges of $10 μC$ each are situated $10 cm$ apart from each other. What is the increase in the electrostatic potential energy of the system when the two charges are brought closer by $2 cm$ ?