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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359541 In uniform electric field $\vec{E} = 10 N / C$ as shown in the figure. The value of $(V_{A} - V_{B})$ is
supporting img

1

10 V

2

- 10 V

3

20 V

4

- 20 V

Explanation:

supporting img
$E = \frac{- Δ V}{Δ r}$
$Δ V = - E . Δ r = 10 \times 2 \times \cos 150^{\circ}$
$= + 10 V$
Since $V_{B} > V_{A}, V_{A} - V_{B}$ will be negative

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359542 The potential at a point $x$ (measured in $μ$ $m$ ) due to some charges situated on the $x$ -axis is given by $V (x) = \frac{20}{x^{2} - 4}$ volt. The electric field $E$ at $x = 4 μ$ $m$ is given by

1

\frac{5}{3} \frac{V}{μ m}

and in the positive

x

- direction

2

\frac{10}{9} \frac{V}{μ m}

and in the negative

x

- direction

3

\frac{10}{9} \frac{V}{μ m}

and in the positive

x

-direction

4

\frac{5}{3} \frac{V}{μ m}

and in the negative direction

Explanation:

The $x$ - component of electric field is
$E_{x} = \frac{- d V}{d x} = - 20 \frac{d}{d x} (\frac{1}{x^{2} - 4})$
$E_{x} = \frac{40 x}{(x^{2} - 4)} \Rightarrow E_{x} = \frac{10}{9} \frac{V}{μ m} (at x = 4 μ m)$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359543 The electric potential $V$ at any point $(x, y, z)$ , all in metres in space is given by $V = 4 x^{2}$ volt. The electric field at the point (1, 0, 2) in volt/meter, is

1 8 along positive

X

-axis

2 8 along negative

X

-axis

3 16 along negative

X

-axis

4 16 along positive

X

-axis

Explanation:

$\vec{E} = - \frac{d V}{d x} \hat{i} = - 8 x \hat{i}$ volt/meter
$E_{(1, 0, 2)} = - 8 i V / m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359544 The electric field $\vec{E}$ between two points is constant in both magnitude and direction. Consider a path of length $d$ at an angle $θ = 60^{\circ}$ with respect to field lines shown in the figure. The potential difference between points 1 and 2 is
supporting img

1

\frac{E}{d \cos 60^{\circ}}

2

E d \cos 60^{\circ}

3

\frac{E d}{\cos 60^{\circ}}

4

\frac{E}{d} \cos 60^{\circ}

Explanation:

Points 2 and 3 are equipotential points. Hence, potential difference between points 1 and 2 is the same as that between 1 and 3.
$∴ V = E d \cos 60^{\circ}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359541 In uniform electric field $\vec{E} = 10 N / C$ as shown in the figure. The value of $(V_{A} - V_{B})$ is
supporting img

1

10 V

2

- 10 V

3

20 V

4

- 20 V

Explanation:

supporting img
$E = \frac{- Δ V}{Δ r}$
$Δ V = - E . Δ r = 10 \times 2 \times \cos 150^{\circ}$
$= + 10 V$
Since $V_{B} > V_{A}, V_{A} - V_{B}$ will be negative

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359542 The potential at a point $x$ (measured in $μ$ $m$ ) due to some charges situated on the $x$ -axis is given by $V (x) = \frac{20}{x^{2} - 4}$ volt. The electric field $E$ at $x = 4 μ$ $m$ is given by

1

\frac{5}{3} \frac{V}{μ m}

and in the positive

x

- direction

2

\frac{10}{9} \frac{V}{μ m}

and in the negative

x

- direction

3

\frac{10}{9} \frac{V}{μ m}

and in the positive

x

-direction

4

\frac{5}{3} \frac{V}{μ m}

and in the negative direction

Explanation:

The $x$ - component of electric field is
$E_{x} = \frac{- d V}{d x} = - 20 \frac{d}{d x} (\frac{1}{x^{2} - 4})$
$E_{x} = \frac{40 x}{(x^{2} - 4)} \Rightarrow E_{x} = \frac{10}{9} \frac{V}{μ m} (at x = 4 μ m)$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359543 The electric potential $V$ at any point $(x, y, z)$ , all in metres in space is given by $V = 4 x^{2}$ volt. The electric field at the point (1, 0, 2) in volt/meter, is

1 8 along positive

X

-axis

2 8 along negative

X

-axis

3 16 along negative

X

-axis

4 16 along positive

X

-axis

Explanation:

$\vec{E} = - \frac{d V}{d x} \hat{i} = - 8 x \hat{i}$ volt/meter
$E_{(1, 0, 2)} = - 8 i V / m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359544 The electric field $\vec{E}$ between two points is constant in both magnitude and direction. Consider a path of length $d$ at an angle $θ = 60^{\circ}$ with respect to field lines shown in the figure. The potential difference between points 1 and 2 is
supporting img

1

\frac{E}{d \cos 60^{\circ}}

2

E d \cos 60^{\circ}

3

\frac{E d}{\cos 60^{\circ}}

4

\frac{E}{d} \cos 60^{\circ}

Explanation:

Points 2 and 3 are equipotential points. Hence, potential difference between points 1 and 2 is the same as that between 1 and 3.
$∴ V = E d \cos 60^{\circ}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359541 In uniform electric field $\vec{E} = 10 N / C$ as shown in the figure. The value of $(V_{A} - V_{B})$ is
supporting img

1

10 V

2

- 10 V

3

20 V

4

- 20 V

Explanation:

supporting img
$E = \frac{- Δ V}{Δ r}$
$Δ V = - E . Δ r = 10 \times 2 \times \cos 150^{\circ}$
$= + 10 V$
Since $V_{B} > V_{A}, V_{A} - V_{B}$ will be negative

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359542 The potential at a point $x$ (measured in $μ$ $m$ ) due to some charges situated on the $x$ -axis is given by $V (x) = \frac{20}{x^{2} - 4}$ volt. The electric field $E$ at $x = 4 μ$ $m$ is given by

1

\frac{5}{3} \frac{V}{μ m}

and in the positive

x

- direction

2

\frac{10}{9} \frac{V}{μ m}

and in the negative

x

- direction

3

\frac{10}{9} \frac{V}{μ m}

and in the positive

x

-direction

4

\frac{5}{3} \frac{V}{μ m}

and in the negative direction

Explanation:

The $x$ - component of electric field is
$E_{x} = \frac{- d V}{d x} = - 20 \frac{d}{d x} (\frac{1}{x^{2} - 4})$
$E_{x} = \frac{40 x}{(x^{2} - 4)} \Rightarrow E_{x} = \frac{10}{9} \frac{V}{μ m} (at x = 4 μ m)$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359543 The electric potential $V$ at any point $(x, y, z)$ , all in metres in space is given by $V = 4 x^{2}$ volt. The electric field at the point (1, 0, 2) in volt/meter, is

1 8 along positive

X

-axis

2 8 along negative

X

-axis

3 16 along negative

X

-axis

4 16 along positive

X

-axis

Explanation:

$\vec{E} = - \frac{d V}{d x} \hat{i} = - 8 x \hat{i}$ volt/meter
$E_{(1, 0, 2)} = - 8 i V / m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359544 The electric field $\vec{E}$ between two points is constant in both magnitude and direction. Consider a path of length $d$ at an angle $θ = 60^{\circ}$ with respect to field lines shown in the figure. The potential difference between points 1 and 2 is
supporting img

1

\frac{E}{d \cos 60^{\circ}}

2

E d \cos 60^{\circ}

3

\frac{E d}{\cos 60^{\circ}}

4

\frac{E}{d} \cos 60^{\circ}

Explanation:

Points 2 and 3 are equipotential points. Hence, potential difference between points 1 and 2 is the same as that between 1 and 3.
$∴ V = E d \cos 60^{\circ}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359541 In uniform electric field $\vec{E} = 10 N / C$ as shown in the figure. The value of $(V_{A} - V_{B})$ is
supporting img

1

10 V

2

- 10 V

3

20 V

4

- 20 V

Explanation:

supporting img
$E = \frac{- Δ V}{Δ r}$
$Δ V = - E . Δ r = 10 \times 2 \times \cos 150^{\circ}$
$= + 10 V$
Since $V_{B} > V_{A}, V_{A} - V_{B}$ will be negative

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359542 The potential at a point $x$ (measured in $μ$ $m$ ) due to some charges situated on the $x$ -axis is given by $V (x) = \frac{20}{x^{2} - 4}$ volt. The electric field $E$ at $x = 4 μ$ $m$ is given by

1

\frac{5}{3} \frac{V}{μ m}

and in the positive

x

- direction

2

\frac{10}{9} \frac{V}{μ m}

and in the negative

x

- direction

3

\frac{10}{9} \frac{V}{μ m}

and in the positive

x

-direction

4

\frac{5}{3} \frac{V}{μ m}

and in the negative direction

Explanation:

The $x$ - component of electric field is
$E_{x} = \frac{- d V}{d x} = - 20 \frac{d}{d x} (\frac{1}{x^{2} - 4})$
$E_{x} = \frac{40 x}{(x^{2} - 4)} \Rightarrow E_{x} = \frac{10}{9} \frac{V}{μ m} (at x = 4 μ m)$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359543 The electric potential $V$ at any point $(x, y, z)$ , all in metres in space is given by $V = 4 x^{2}$ volt. The electric field at the point (1, 0, 2) in volt/meter, is

1 8 along positive

X

-axis

2 8 along negative

X

-axis

3 16 along negative

X

-axis

4 16 along positive

X

-axis

Explanation:

$\vec{E} = - \frac{d V}{d x} \hat{i} = - 8 x \hat{i}$ volt/meter
$E_{(1, 0, 2)} = - 8 i V / m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359544 The electric field $\vec{E}$ between two points is constant in both magnitude and direction. Consider a path of length $d$ at an angle $θ = 60^{\circ}$ with respect to field lines shown in the figure. The potential difference between points 1 and 2 is
supporting img

1

\frac{E}{d \cos 60^{\circ}}

2

E d \cos 60^{\circ}

3

\frac{E d}{\cos 60^{\circ}}

4

\frac{E}{d} \cos 60^{\circ}

Explanation:

Points 2 and 3 are equipotential points. Hence, potential difference between points 1 and 2 is the same as that between 1 and 3.
$∴ V = E d \cos 60^{\circ}$

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