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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359422 The electrostatic potential of a uniformly charged thin spherical shell of charge $Q$ and radius $R$ at a distance $r$ from the centre is

1

\frac{Q}{4 π ε_{0} r}

for points outside and

\frac{Q}{4 π ε_{0} R}

for points on surface of the sphere

2

\frac{Q}{4 π ε_{0} r}

for both points inside and outside the shell

3 zero for points outside and

\frac{Q}{4 π ε_{0} r}

for points inside the shel

4 zero for both points inside and outside the shell

Explanation:

If charge on a conducting sphere of radius $R$ is $Q$ , then potential outside the sphere, $V_{out} = \frac{1}{4 π ε_{0}} \frac{Q}{r}$
At the surface of sphere, $V_{s} = \frac{1}{4 π ε_{0}} \frac{Q}{R}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359423 An infinite plane of charge with $σ = 2 ϵ_{0} \frac{C}{m^{2}}$ is tilted at a $37^{\circ}$ angle to the vertical direction as shown below. Find the potential difference, $V_{A} - V_{B}$ in volts, between points $A$ and $B$ at 5 $m$ distance apart (where $B$ is vertically above $A) .$
supporting img

1

1 V

2

5 V

3

7 V

4

3 V

Explanation:

supporting img
$E = \frac{σ}{2 ε_{0}} = 1 N / C$
$∴ V_{B} - V_{A} = - \int \vec{E} \cdot d \vec{l}$
$= (\frac{σ}{2_{0}}) 5 \cos 53^{\circ} = 3 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359424 A charge $Q$ is uniformly distributed over a long rod $A B$ of length $L$ as shown in the figure. The electric potential at the point $O$ lying at distance $L$ from the end $A$ is :
supporting img

1

\frac{Q}{8 π ε_{0} L}

2

\frac{3 Q}{4 π ε_{0} L}

3

\frac{3 Q}{4 π ε_{0} L \ln 2}

4

\frac{Q I n 2}{4 π ε_{0} L}

Explanation:

supporting img
$V = \int_{L}^{2 L} \frac{1}{4 π ε_{0}} \frac{d q}{x} = \int_{L}^{2 L} \frac{1}{4 π ε_{0}} \frac{(\frac{Q}{L}) d x}{x} = \frac{Q}{4 π ε_{0} L} \ln (2)$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359425 A unit positive point charge of mass m is projected with a velocity $v$ inside the tunnel which is made inside a uniformly charged non-conducting sphere of charge density $ρ .$ The minimum velocity with which the point charge should be projected such that it can reach the opposite end of the tunnel, is equal to:
supporting img

1

[ρ R^{2} / 4 m ε_{0}]^{1 / 2}

2

[ρ R^{2} / 6 m ε_{0}]^{1 / 2}

3

[ρ R^{2} / 24 m ε_{0}]^{1 / 2}

4 Zero because the initial and the final points are at same potentials

Explanation:

The positive charge will be repelled by the positively charged sphere. It should be projected with initial kinetic energy which will be converted into potential energy. Here the particle should be able to reach the centre of the tunnel. After that it will automatically reach the other end.
$K . E_{i} - K . E_{f} = q (V_{f} - V_{i})$
$K . E_{i} = \frac{1}{2} m v^{2}, V_{i} = \frac{Q}{4 π ε_{0} R} = \frac{σ R^{2}}{3 ε_{0}}$
$V_{f} = \frac{1}{4 π ε_{0}} \frac{Q}{2 R} (3 - \frac{r^{2}}{R^{2}}) (r = \frac{R}{2})$
$\Rightarrow V_{f} = \frac{11 Q}{32 π ε_{0} R} = \frac{11 R^{2} ρ}{24 ε_{0}} \sum_{i = 1}^{n} X_{i}$
Where $ρ = \frac{Q}{\frac{4}{3} π R^{3}}$
$\frac{1}{2} m v^{2} = \frac{11 R^{2} ρ}{24 ε_{0}} - \frac{ρ R^{2}}{3 ε_{0}}$
$v = \sqrt{\frac{ρ R^{2}}{4 m ε_{0}}}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359422 The electrostatic potential of a uniformly charged thin spherical shell of charge $Q$ and radius $R$ at a distance $r$ from the centre is

1

\frac{Q}{4 π ε_{0} r}

for points outside and

\frac{Q}{4 π ε_{0} R}

for points on surface of the sphere

2

\frac{Q}{4 π ε_{0} r}

for both points inside and outside the shell

3 zero for points outside and

\frac{Q}{4 π ε_{0} r}

for points inside the shel

4 zero for both points inside and outside the shell

Explanation:

If charge on a conducting sphere of radius $R$ is $Q$ , then potential outside the sphere, $V_{out} = \frac{1}{4 π ε_{0}} \frac{Q}{r}$
At the surface of sphere, $V_{s} = \frac{1}{4 π ε_{0}} \frac{Q}{R}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359423 An infinite plane of charge with $σ = 2 ϵ_{0} \frac{C}{m^{2}}$ is tilted at a $37^{\circ}$ angle to the vertical direction as shown below. Find the potential difference, $V_{A} - V_{B}$ in volts, between points $A$ and $B$ at 5 $m$ distance apart (where $B$ is vertically above $A) .$
supporting img

1

1 V

2

5 V

3

7 V

4

3 V

Explanation:

supporting img
$E = \frac{σ}{2 ε_{0}} = 1 N / C$
$∴ V_{B} - V_{A} = - \int \vec{E} \cdot d \vec{l}$
$= (\frac{σ}{2_{0}}) 5 \cos 53^{\circ} = 3 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359424 A charge $Q$ is uniformly distributed over a long rod $A B$ of length $L$ as shown in the figure. The electric potential at the point $O$ lying at distance $L$ from the end $A$ is :
supporting img

1

\frac{Q}{8 π ε_{0} L}

2

\frac{3 Q}{4 π ε_{0} L}

3

\frac{3 Q}{4 π ε_{0} L \ln 2}

4

\frac{Q I n 2}{4 π ε_{0} L}

Explanation:

supporting img
$V = \int_{L}^{2 L} \frac{1}{4 π ε_{0}} \frac{d q}{x} = \int_{L}^{2 L} \frac{1}{4 π ε_{0}} \frac{(\frac{Q}{L}) d x}{x} = \frac{Q}{4 π ε_{0} L} \ln (2)$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359425 A unit positive point charge of mass m is projected with a velocity $v$ inside the tunnel which is made inside a uniformly charged non-conducting sphere of charge density $ρ .$ The minimum velocity with which the point charge should be projected such that it can reach the opposite end of the tunnel, is equal to:
supporting img

1

[ρ R^{2} / 4 m ε_{0}]^{1 / 2}

2

[ρ R^{2} / 6 m ε_{0}]^{1 / 2}

3

[ρ R^{2} / 24 m ε_{0}]^{1 / 2}

4 Zero because the initial and the final points are at same potentials

Explanation:

The positive charge will be repelled by the positively charged sphere. It should be projected with initial kinetic energy which will be converted into potential energy. Here the particle should be able to reach the centre of the tunnel. After that it will automatically reach the other end.
$K . E_{i} - K . E_{f} = q (V_{f} - V_{i})$
$K . E_{i} = \frac{1}{2} m v^{2}, V_{i} = \frac{Q}{4 π ε_{0} R} = \frac{σ R^{2}}{3 ε_{0}}$
$V_{f} = \frac{1}{4 π ε_{0}} \frac{Q}{2 R} (3 - \frac{r^{2}}{R^{2}}) (r = \frac{R}{2})$
$\Rightarrow V_{f} = \frac{11 Q}{32 π ε_{0} R} = \frac{11 R^{2} ρ}{24 ε_{0}} \sum_{i = 1}^{n} X_{i}$
Where $ρ = \frac{Q}{\frac{4}{3} π R^{3}}$
$\frac{1}{2} m v^{2} = \frac{11 R^{2} ρ}{24 ε_{0}} - \frac{ρ R^{2}}{3 ε_{0}}$
$v = \sqrt{\frac{ρ R^{2}}{4 m ε_{0}}}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359422 The electrostatic potential of a uniformly charged thin spherical shell of charge $Q$ and radius $R$ at a distance $r$ from the centre is

1

\frac{Q}{4 π ε_{0} r}

for points outside and

\frac{Q}{4 π ε_{0} R}

for points on surface of the sphere

2

\frac{Q}{4 π ε_{0} r}

for both points inside and outside the shell

3 zero for points outside and

\frac{Q}{4 π ε_{0} r}

for points inside the shel

4 zero for both points inside and outside the shell

Explanation:

If charge on a conducting sphere of radius $R$ is $Q$ , then potential outside the sphere, $V_{out} = \frac{1}{4 π ε_{0}} \frac{Q}{r}$
At the surface of sphere, $V_{s} = \frac{1}{4 π ε_{0}} \frac{Q}{R}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359423 An infinite plane of charge with $σ = 2 ϵ_{0} \frac{C}{m^{2}}$ is tilted at a $37^{\circ}$ angle to the vertical direction as shown below. Find the potential difference, $V_{A} - V_{B}$ in volts, between points $A$ and $B$ at 5 $m$ distance apart (where $B$ is vertically above $A) .$
supporting img

1

1 V

2

5 V

3

7 V

4

3 V

Explanation:

supporting img
$E = \frac{σ}{2 ε_{0}} = 1 N / C$
$∴ V_{B} - V_{A} = - \int \vec{E} \cdot d \vec{l}$
$= (\frac{σ}{2_{0}}) 5 \cos 53^{\circ} = 3 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359424 A charge $Q$ is uniformly distributed over a long rod $A B$ of length $L$ as shown in the figure. The electric potential at the point $O$ lying at distance $L$ from the end $A$ is :
supporting img

1

\frac{Q}{8 π ε_{0} L}

2

\frac{3 Q}{4 π ε_{0} L}

3

\frac{3 Q}{4 π ε_{0} L \ln 2}

4

\frac{Q I n 2}{4 π ε_{0} L}

Explanation:

supporting img
$V = \int_{L}^{2 L} \frac{1}{4 π ε_{0}} \frac{d q}{x} = \int_{L}^{2 L} \frac{1}{4 π ε_{0}} \frac{(\frac{Q}{L}) d x}{x} = \frac{Q}{4 π ε_{0} L} \ln (2)$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359425 A unit positive point charge of mass m is projected with a velocity $v$ inside the tunnel which is made inside a uniformly charged non-conducting sphere of charge density $ρ .$ The minimum velocity with which the point charge should be projected such that it can reach the opposite end of the tunnel, is equal to:
supporting img

1

[ρ R^{2} / 4 m ε_{0}]^{1 / 2}

2

[ρ R^{2} / 6 m ε_{0}]^{1 / 2}

3

[ρ R^{2} / 24 m ε_{0}]^{1 / 2}

4 Zero because the initial and the final points are at same potentials

Explanation:

The positive charge will be repelled by the positively charged sphere. It should be projected with initial kinetic energy which will be converted into potential energy. Here the particle should be able to reach the centre of the tunnel. After that it will automatically reach the other end.
$K . E_{i} - K . E_{f} = q (V_{f} - V_{i})$
$K . E_{i} = \frac{1}{2} m v^{2}, V_{i} = \frac{Q}{4 π ε_{0} R} = \frac{σ R^{2}}{3 ε_{0}}$
$V_{f} = \frac{1}{4 π ε_{0}} \frac{Q}{2 R} (3 - \frac{r^{2}}{R^{2}}) (r = \frac{R}{2})$
$\Rightarrow V_{f} = \frac{11 Q}{32 π ε_{0} R} = \frac{11 R^{2} ρ}{24 ε_{0}} \sum_{i = 1}^{n} X_{i}$
Where $ρ = \frac{Q}{\frac{4}{3} π R^{3}}$
$\frac{1}{2} m v^{2} = \frac{11 R^{2} ρ}{24 ε_{0}} - \frac{ρ R^{2}}{3 ε_{0}}$
$v = \sqrt{\frac{ρ R^{2}}{4 m ε_{0}}}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359422 The electrostatic potential of a uniformly charged thin spherical shell of charge $Q$ and radius $R$ at a distance $r$ from the centre is

1

\frac{Q}{4 π ε_{0} r}

for points outside and

\frac{Q}{4 π ε_{0} R}

for points on surface of the sphere

2

\frac{Q}{4 π ε_{0} r}

for both points inside and outside the shell

3 zero for points outside and

\frac{Q}{4 π ε_{0} r}

for points inside the shel

4 zero for both points inside and outside the shell

Explanation:

If charge on a conducting sphere of radius $R$ is $Q$ , then potential outside the sphere, $V_{out} = \frac{1}{4 π ε_{0}} \frac{Q}{r}$
At the surface of sphere, $V_{s} = \frac{1}{4 π ε_{0}} \frac{Q}{R}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359423 An infinite plane of charge with $σ = 2 ϵ_{0} \frac{C}{m^{2}}$ is tilted at a $37^{\circ}$ angle to the vertical direction as shown below. Find the potential difference, $V_{A} - V_{B}$ in volts, between points $A$ and $B$ at 5 $m$ distance apart (where $B$ is vertically above $A) .$
supporting img

1

1 V

2

5 V

3

7 V

4

3 V

Explanation:

supporting img
$E = \frac{σ}{2 ε_{0}} = 1 N / C$
$∴ V_{B} - V_{A} = - \int \vec{E} \cdot d \vec{l}$
$= (\frac{σ}{2_{0}}) 5 \cos 53^{\circ} = 3 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359424 A charge $Q$ is uniformly distributed over a long rod $A B$ of length $L$ as shown in the figure. The electric potential at the point $O$ lying at distance $L$ from the end $A$ is :
supporting img

1

\frac{Q}{8 π ε_{0} L}

2

\frac{3 Q}{4 π ε_{0} L}

3

\frac{3 Q}{4 π ε_{0} L \ln 2}

4

\frac{Q I n 2}{4 π ε_{0} L}

Explanation:

supporting img
$V = \int_{L}^{2 L} \frac{1}{4 π ε_{0}} \frac{d q}{x} = \int_{L}^{2 L} \frac{1}{4 π ε_{0}} \frac{(\frac{Q}{L}) d x}{x} = \frac{Q}{4 π ε_{0} L} \ln (2)$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359425 A unit positive point charge of mass m is projected with a velocity $v$ inside the tunnel which is made inside a uniformly charged non-conducting sphere of charge density $ρ .$ The minimum velocity with which the point charge should be projected such that it can reach the opposite end of the tunnel, is equal to:
supporting img

1

[ρ R^{2} / 4 m ε_{0}]^{1 / 2}

2

[ρ R^{2} / 6 m ε_{0}]^{1 / 2}

3

[ρ R^{2} / 24 m ε_{0}]^{1 / 2}

4 Zero because the initial and the final points are at same potentials

Explanation:

The positive charge will be repelled by the positively charged sphere. It should be projected with initial kinetic energy which will be converted into potential energy. Here the particle should be able to reach the centre of the tunnel. After that it will automatically reach the other end.
$K . E_{i} - K . E_{f} = q (V_{f} - V_{i})$
$K . E_{i} = \frac{1}{2} m v^{2}, V_{i} = \frac{Q}{4 π ε_{0} R} = \frac{σ R^{2}}{3 ε_{0}}$
$V_{f} = \frac{1}{4 π ε_{0}} \frac{Q}{2 R} (3 - \frac{r^{2}}{R^{2}}) (r = \frac{R}{2})$
$\Rightarrow V_{f} = \frac{11 Q}{32 π ε_{0} R} = \frac{11 R^{2} ρ}{24 ε_{0}} \sum_{i = 1}^{n} X_{i}$
Where $ρ = \frac{Q}{\frac{4}{3} π R^{3}}$
$\frac{1}{2} m v^{2} = \frac{11 R^{2} ρ}{24 ε_{0}} - \frac{ρ R^{2}}{3 ε_{0}}$
$v = \sqrt{\frac{ρ R^{2}}{4 m ε_{0}}}$

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