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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359345 A dipole is rotated in an external field E the plane of paper from angle $θ_{1}$ to an angle $θ_{2}$ . The amount of work done by the external agent will be given by ( $θ$ is the angle between $\vec{p}$ and $\vec{E}$ )

1

p E (\cos θ_{2} - \cos θ_{1})

2

- 2 p E (\cos θ_{1} - \cos θ_{2})

3

p E (\cos θ_{1} - \cos θ_{2})

4

- 2 p E (\cos θ_{1} - \cos θ_{2})

Explanation:

The amount of work done by the external agent is given by
$W = Δ U = U_{2} - U_{1}$
$= - p E \cos θ_{2} - (- p E \cos θ_{1})$
$= p E (\cos θ_{1} - \cos θ_{2})$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359346 An electric dipole of moment $\vec{p}$ is placed normal to the lines of force of electric intensity $\vec{E}$ , then the work done in deflecting it through an angle of $180^{\circ}$ is

1

p E

2

+ 2 p E

3

- 2 p E

4

Z e r o

Explanation:

Work done by an external agent is
$W = U_{2} - U_{1}$
$U_{1} = P E \cos 90^{\circ} \Rightarrow U_{2} = P E \cos (180^{\circ} + 90^{\circ})$
$W = 0$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359347 An electric dipole, made up of positive and negative charges each of $30 μ C$ and placed at a distance 10 $c m$ apart is placed in an electric field of $5 \times 10^{5} N C^{- 1}$ . Find the work done to turn the dipole from $0^{\circ}$ to $180^{\circ}$

1

3 J

2

15 J

3

7 J

4

25 J

Explanation:

Work done in rotating the dipole from an angle $θ_{1} t o θ_{2}$ is
$W = Δ U = U_{2} - U_{1} = - p E \cos θ_{2} + p E \cos θ_{1}$
$W = p E (\cos θ_{1} - \cos θ_{2})$
$W = \vec{p} E (\cos 0^{\circ} - \cos 180^{\circ})$
where, $W = p E (\cos 0^{\circ} - \cos 180^{\circ})$
$\Rightarrow W = 2 p E = 2 \times q \times d \times E$
$= 2 \times (30 \times 10^{- 6}) \times (10 \times 10^{- 2}) \times 5 \times 10^{5}$
$= 3 J$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359348 An electric dipole of length 1 $c m$ is placed with the axis making an angle of $30^{\circ}$ to an electric field of strength $10^{3} N / C$ . If it experiances a torque of $10 \sqrt{3} N m$ , the potential energy of the dipole is

1

50 J

2

12 J

3

30 J

4

250 J

Explanation:

As, $τ = p E \sin θ$
Given that
$\Rightarrow 10 \sqrt{3} = p \times 10^{3} \sin 30^{\circ}$
As, $p = 2 \sqrt{3} \times 10^{- 2} c m$
$U = p E \cos θ$
$U = 2 \sqrt{3} \times 10^{- 2} \times 10^{3} \cos 30^{\circ}$
$∴$ Potential energy $U = 30 J$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359345 A dipole is rotated in an external field E the plane of paper from angle $θ_{1}$ to an angle $θ_{2}$ . The amount of work done by the external agent will be given by ( $θ$ is the angle between $\vec{p}$ and $\vec{E}$ )

1

p E (\cos θ_{2} - \cos θ_{1})

2

- 2 p E (\cos θ_{1} - \cos θ_{2})

3

p E (\cos θ_{1} - \cos θ_{2})

4

- 2 p E (\cos θ_{1} - \cos θ_{2})

Explanation:

The amount of work done by the external agent is given by
$W = Δ U = U_{2} - U_{1}$
$= - p E \cos θ_{2} - (- p E \cos θ_{1})$
$= p E (\cos θ_{1} - \cos θ_{2})$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359346 An electric dipole of moment $\vec{p}$ is placed normal to the lines of force of electric intensity $\vec{E}$ , then the work done in deflecting it through an angle of $180^{\circ}$ is

1

p E

2

+ 2 p E

3

- 2 p E

4

Z e r o

Explanation:

Work done by an external agent is
$W = U_{2} - U_{1}$
$U_{1} = P E \cos 90^{\circ} \Rightarrow U_{2} = P E \cos (180^{\circ} + 90^{\circ})$
$W = 0$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359347 An electric dipole, made up of positive and negative charges each of $30 μ C$ and placed at a distance 10 $c m$ apart is placed in an electric field of $5 \times 10^{5} N C^{- 1}$ . Find the work done to turn the dipole from $0^{\circ}$ to $180^{\circ}$

1

3 J

2

15 J

3

7 J

4

25 J

Explanation:

Work done in rotating the dipole from an angle $θ_{1} t o θ_{2}$ is
$W = Δ U = U_{2} - U_{1} = - p E \cos θ_{2} + p E \cos θ_{1}$
$W = p E (\cos θ_{1} - \cos θ_{2})$
$W = \vec{p} E (\cos 0^{\circ} - \cos 180^{\circ})$
where, $W = p E (\cos 0^{\circ} - \cos 180^{\circ})$
$\Rightarrow W = 2 p E = 2 \times q \times d \times E$
$= 2 \times (30 \times 10^{- 6}) \times (10 \times 10^{- 2}) \times 5 \times 10^{5}$
$= 3 J$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359348 An electric dipole of length 1 $c m$ is placed with the axis making an angle of $30^{\circ}$ to an electric field of strength $10^{3} N / C$ . If it experiances a torque of $10 \sqrt{3} N m$ , the potential energy of the dipole is

1

50 J

2

12 J

3

30 J

4

250 J

Explanation:

As, $τ = p E \sin θ$
Given that
$\Rightarrow 10 \sqrt{3} = p \times 10^{3} \sin 30^{\circ}$
As, $p = 2 \sqrt{3} \times 10^{- 2} c m$
$U = p E \cos θ$
$U = 2 \sqrt{3} \times 10^{- 2} \times 10^{3} \cos 30^{\circ}$
$∴$ Potential energy $U = 30 J$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359345 A dipole is rotated in an external field E the plane of paper from angle $θ_{1}$ to an angle $θ_{2}$ . The amount of work done by the external agent will be given by ( $θ$ is the angle between $\vec{p}$ and $\vec{E}$ )

1

p E (\cos θ_{2} - \cos θ_{1})

2

- 2 p E (\cos θ_{1} - \cos θ_{2})

3

p E (\cos θ_{1} - \cos θ_{2})

4

- 2 p E (\cos θ_{1} - \cos θ_{2})

Explanation:

The amount of work done by the external agent is given by
$W = Δ U = U_{2} - U_{1}$
$= - p E \cos θ_{2} - (- p E \cos θ_{1})$
$= p E (\cos θ_{1} - \cos θ_{2})$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359346 An electric dipole of moment $\vec{p}$ is placed normal to the lines of force of electric intensity $\vec{E}$ , then the work done in deflecting it through an angle of $180^{\circ}$ is

1

p E

2

+ 2 p E

3

- 2 p E

4

Z e r o

Explanation:

Work done by an external agent is
$W = U_{2} - U_{1}$
$U_{1} = P E \cos 90^{\circ} \Rightarrow U_{2} = P E \cos (180^{\circ} + 90^{\circ})$
$W = 0$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359347 An electric dipole, made up of positive and negative charges each of $30 μ C$ and placed at a distance 10 $c m$ apart is placed in an electric field of $5 \times 10^{5} N C^{- 1}$ . Find the work done to turn the dipole from $0^{\circ}$ to $180^{\circ}$

1

3 J

2

15 J

3

7 J

4

25 J

Explanation:

Work done in rotating the dipole from an angle $θ_{1} t o θ_{2}$ is
$W = Δ U = U_{2} - U_{1} = - p E \cos θ_{2} + p E \cos θ_{1}$
$W = p E (\cos θ_{1} - \cos θ_{2})$
$W = \vec{p} E (\cos 0^{\circ} - \cos 180^{\circ})$
where, $W = p E (\cos 0^{\circ} - \cos 180^{\circ})$
$\Rightarrow W = 2 p E = 2 \times q \times d \times E$
$= 2 \times (30 \times 10^{- 6}) \times (10 \times 10^{- 2}) \times 5 \times 10^{5}$
$= 3 J$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359348 An electric dipole of length 1 $c m$ is placed with the axis making an angle of $30^{\circ}$ to an electric field of strength $10^{3} N / C$ . If it experiances a torque of $10 \sqrt{3} N m$ , the potential energy of the dipole is

1

50 J

2

12 J

3

30 J

4

250 J

Explanation:

As, $τ = p E \sin θ$
Given that
$\Rightarrow 10 \sqrt{3} = p \times 10^{3} \sin 30^{\circ}$
As, $p = 2 \sqrt{3} \times 10^{- 2} c m$
$U = p E \cos θ$
$U = 2 \sqrt{3} \times 10^{- 2} \times 10^{3} \cos 30^{\circ}$
$∴$ Potential energy $U = 30 J$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359345 A dipole is rotated in an external field E the plane of paper from angle $θ_{1}$ to an angle $θ_{2}$ . The amount of work done by the external agent will be given by ( $θ$ is the angle between $\vec{p}$ and $\vec{E}$ )

1

p E (\cos θ_{2} - \cos θ_{1})

2

- 2 p E (\cos θ_{1} - \cos θ_{2})

3

p E (\cos θ_{1} - \cos θ_{2})

4

- 2 p E (\cos θ_{1} - \cos θ_{2})

Explanation:

The amount of work done by the external agent is given by
$W = Δ U = U_{2} - U_{1}$
$= - p E \cos θ_{2} - (- p E \cos θ_{1})$
$= p E (\cos θ_{1} - \cos θ_{2})$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359346 An electric dipole of moment $\vec{p}$ is placed normal to the lines of force of electric intensity $\vec{E}$ , then the work done in deflecting it through an angle of $180^{\circ}$ is

1

p E

2

+ 2 p E

3

- 2 p E

4

Z e r o

Explanation:

Work done by an external agent is
$W = U_{2} - U_{1}$
$U_{1} = P E \cos 90^{\circ} \Rightarrow U_{2} = P E \cos (180^{\circ} + 90^{\circ})$
$W = 0$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359347 An electric dipole, made up of positive and negative charges each of $30 μ C$ and placed at a distance 10 $c m$ apart is placed in an electric field of $5 \times 10^{5} N C^{- 1}$ . Find the work done to turn the dipole from $0^{\circ}$ to $180^{\circ}$

1

3 J

2

15 J

3

7 J

4

25 J

Explanation:

Work done in rotating the dipole from an angle $θ_{1} t o θ_{2}$ is
$W = Δ U = U_{2} - U_{1} = - p E \cos θ_{2} + p E \cos θ_{1}$
$W = p E (\cos θ_{1} - \cos θ_{2})$
$W = \vec{p} E (\cos 0^{\circ} - \cos 180^{\circ})$
where, $W = p E (\cos 0^{\circ} - \cos 180^{\circ})$
$\Rightarrow W = 2 p E = 2 \times q \times d \times E$
$= 2 \times (30 \times 10^{- 6}) \times (10 \times 10^{- 2}) \times 5 \times 10^{5}$
$= 3 J$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359348 An electric dipole of length 1 $c m$ is placed with the axis making an angle of $30^{\circ}$ to an electric field of strength $10^{3} N / C$ . If it experiances a torque of $10 \sqrt{3} N m$ , the potential energy of the dipole is

1

50 J

2

12 J

3

30 J

4

250 J

Explanation:

As, $τ = p E \sin θ$
Given that
$\Rightarrow 10 \sqrt{3} = p \times 10^{3} \sin 30^{\circ}$
As, $p = 2 \sqrt{3} \times 10^{- 2} c m$
$U = p E \cos θ$
$U = 2 \sqrt{3} \times 10^{- 2} \times 10^{3} \cos 30^{\circ}$
$∴$ Potential energy $U = 30 J$