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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359567 The electric potential at any point $x, y, z$ in metres is given by $V = 3 x^{2}$ . The electric field at a point $(2 m, 0, 1 m)$ is

1

12 V m^{- 1}

2

- 6 V m^{- 1}

3

6 V m^{- 1}

4

- 12 V m^{- 1}

Explanation:

Here, electric potential $V = 3 x^{2}$
Since potential is independent of $y$ and $z$ , electric field components in those two directions are zero. Hence
$E = - \frac{d V}{d x} = - \frac{d}{d x} (3 x^{2}) = - 6 x$
$E_{(2 m .0 m .1 m)} = - 12 V m^{- 1}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359568 A 5 coulomb charge experiences a constant force of $2000 N$ when moved between two points, separated by a distance of $2 c m$ , in a uniform electric field. The potential difference between these two points is

1

8 V

2

200 V

3

800 V

4

20, 000 V

Explanation:

$E = \frac{F}{q}$
$E = \frac{2000}{5} = 400 N / C$
Potential difference,
$Δ V = E \cdot d = 400 \times \frac{2}{100} = 8 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359569 The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11} V .$ The electric field at the point is

1

4 π ε_{0} Q \times 10^{22} V / m

2

12 π ε_{0} Q \times 10^{22} V / m

3

4 π ε_{0} Q \times 10^{20} V / m

4

12 π ε_{0} Q \times 10^{20} V / m

Explanation:

As potential at a point due to a point charge is
$V = \frac{1}{4 π ε_{0}} \times \frac{Q}{r}$
The electric field is
$E = \frac{1}{4 π ε_{0}} \times \frac{Q}{r^{2}}$
Given that $V = Q \times 10^{11}$
From the above equations
$E = 4 π ε_{o} Q \times 10^{22} V m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359570 The electric potential at a point $(x, y, z)$ is given by $V = - x^{2} y - x z^{3} + 4$ The electric field $\vec{E}$ at that point is

1

\vec{E} = \hat{i} (2 x y + z^{3}) + \hat{j} x^{2} + \hat{k} 3 x z^{2}

2

\vec{E} = \hat{i} (2 x y - z^{3}) + \hat{j} x y^{2} + \hat{k} 3 z^{2} x

3

\vec{E} = \hat{i} z^{3} + \hat{j} x y z + \hat{k} z^{2}

4

\vec{E} = \hat{i} 2 x y + \hat{j} (x^{2} + y^{2}) + \hat{k} (3 x z - y^{2})

Explanation:

$E_{x} = - \frac{d V}{d x} = - [- 2 x y - z^{3}]$
$E_{y} = - \frac{d V}{d y} = - [- x^{2}]$
$E_{z} = - \frac{d V}{d z} = - [- 3 x z^{2}]$
$∴ \vec{E} = E_{x} \hat{i} + E_{y} \hat{j} + E_{z} \hat{k} = (2 x y + z^{3}) \hat{i} + x^{2} \hat{j} + 3 x z^{2} \hat{k}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359567 The electric potential at any point $x, y, z$ in metres is given by $V = 3 x^{2}$ . The electric field at a point $(2 m, 0, 1 m)$ is

1

12 V m^{- 1}

2

- 6 V m^{- 1}

3

6 V m^{- 1}

4

- 12 V m^{- 1}

Explanation:

Here, electric potential $V = 3 x^{2}$
Since potential is independent of $y$ and $z$ , electric field components in those two directions are zero. Hence
$E = - \frac{d V}{d x} = - \frac{d}{d x} (3 x^{2}) = - 6 x$
$E_{(2 m .0 m .1 m)} = - 12 V m^{- 1}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359568 A 5 coulomb charge experiences a constant force of $2000 N$ when moved between two points, separated by a distance of $2 c m$ , in a uniform electric field. The potential difference between these two points is

1

8 V

2

200 V

3

800 V

4

20, 000 V

Explanation:

$E = \frac{F}{q}$
$E = \frac{2000}{5} = 400 N / C$
Potential difference,
$Δ V = E \cdot d = 400 \times \frac{2}{100} = 8 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359569 The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11} V .$ The electric field at the point is

1

4 π ε_{0} Q \times 10^{22} V / m

2

12 π ε_{0} Q \times 10^{22} V / m

3

4 π ε_{0} Q \times 10^{20} V / m

4

12 π ε_{0} Q \times 10^{20} V / m

Explanation:

As potential at a point due to a point charge is
$V = \frac{1}{4 π ε_{0}} \times \frac{Q}{r}$
The electric field is
$E = \frac{1}{4 π ε_{0}} \times \frac{Q}{r^{2}}$
Given that $V = Q \times 10^{11}$
From the above equations
$E = 4 π ε_{o} Q \times 10^{22} V m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359570 The electric potential at a point $(x, y, z)$ is given by $V = - x^{2} y - x z^{3} + 4$ The electric field $\vec{E}$ at that point is

1

\vec{E} = \hat{i} (2 x y + z^{3}) + \hat{j} x^{2} + \hat{k} 3 x z^{2}

2

\vec{E} = \hat{i} (2 x y - z^{3}) + \hat{j} x y^{2} + \hat{k} 3 z^{2} x

3

\vec{E} = \hat{i} z^{3} + \hat{j} x y z + \hat{k} z^{2}

4

\vec{E} = \hat{i} 2 x y + \hat{j} (x^{2} + y^{2}) + \hat{k} (3 x z - y^{2})

Explanation:

$E_{x} = - \frac{d V}{d x} = - [- 2 x y - z^{3}]$
$E_{y} = - \frac{d V}{d y} = - [- x^{2}]$
$E_{z} = - \frac{d V}{d z} = - [- 3 x z^{2}]$
$∴ \vec{E} = E_{x} \hat{i} + E_{y} \hat{j} + E_{z} \hat{k} = (2 x y + z^{3}) \hat{i} + x^{2} \hat{j} + 3 x z^{2} \hat{k}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359567 The electric potential at any point $x, y, z$ in metres is given by $V = 3 x^{2}$ . The electric field at a point $(2 m, 0, 1 m)$ is

1

12 V m^{- 1}

2

- 6 V m^{- 1}

3

6 V m^{- 1}

4

- 12 V m^{- 1}

Explanation:

Here, electric potential $V = 3 x^{2}$
Since potential is independent of $y$ and $z$ , electric field components in those two directions are zero. Hence
$E = - \frac{d V}{d x} = - \frac{d}{d x} (3 x^{2}) = - 6 x$
$E_{(2 m .0 m .1 m)} = - 12 V m^{- 1}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359568 A 5 coulomb charge experiences a constant force of $2000 N$ when moved between two points, separated by a distance of $2 c m$ , in a uniform electric field. The potential difference between these two points is

1

8 V

2

200 V

3

800 V

4

20, 000 V

Explanation:

$E = \frac{F}{q}$
$E = \frac{2000}{5} = 400 N / C$
Potential difference,
$Δ V = E \cdot d = 400 \times \frac{2}{100} = 8 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359569 The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11} V .$ The electric field at the point is

1

4 π ε_{0} Q \times 10^{22} V / m

2

12 π ε_{0} Q \times 10^{22} V / m

3

4 π ε_{0} Q \times 10^{20} V / m

4

12 π ε_{0} Q \times 10^{20} V / m

Explanation:

As potential at a point due to a point charge is
$V = \frac{1}{4 π ε_{0}} \times \frac{Q}{r}$
The electric field is
$E = \frac{1}{4 π ε_{0}} \times \frac{Q}{r^{2}}$
Given that $V = Q \times 10^{11}$
From the above equations
$E = 4 π ε_{o} Q \times 10^{22} V m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359570 The electric potential at a point $(x, y, z)$ is given by $V = - x^{2} y - x z^{3} + 4$ The electric field $\vec{E}$ at that point is

1

\vec{E} = \hat{i} (2 x y + z^{3}) + \hat{j} x^{2} + \hat{k} 3 x z^{2}

2

\vec{E} = \hat{i} (2 x y - z^{3}) + \hat{j} x y^{2} + \hat{k} 3 z^{2} x

3

\vec{E} = \hat{i} z^{3} + \hat{j} x y z + \hat{k} z^{2}

4

\vec{E} = \hat{i} 2 x y + \hat{j} (x^{2} + y^{2}) + \hat{k} (3 x z - y^{2})

Explanation:

$E_{x} = - \frac{d V}{d x} = - [- 2 x y - z^{3}]$
$E_{y} = - \frac{d V}{d y} = - [- x^{2}]$
$E_{z} = - \frac{d V}{d z} = - [- 3 x z^{2}]$
$∴ \vec{E} = E_{x} \hat{i} + E_{y} \hat{j} + E_{z} \hat{k} = (2 x y + z^{3}) \hat{i} + x^{2} \hat{j} + 3 x z^{2} \hat{k}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359567 The electric potential at any point $x, y, z$ in metres is given by $V = 3 x^{2}$ . The electric field at a point $(2 m, 0, 1 m)$ is

1

12 V m^{- 1}

2

- 6 V m^{- 1}

3

6 V m^{- 1}

4

- 12 V m^{- 1}

Explanation:

Here, electric potential $V = 3 x^{2}$
Since potential is independent of $y$ and $z$ , electric field components in those two directions are zero. Hence
$E = - \frac{d V}{d x} = - \frac{d}{d x} (3 x^{2}) = - 6 x$
$E_{(2 m .0 m .1 m)} = - 12 V m^{- 1}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359568 A 5 coulomb charge experiences a constant force of $2000 N$ when moved between two points, separated by a distance of $2 c m$ , in a uniform electric field. The potential difference between these two points is

1

8 V

2

200 V

3

800 V

4

20, 000 V

Explanation:

$E = \frac{F}{q}$
$E = \frac{2000}{5} = 400 N / C$
Potential difference,
$Δ V = E \cdot d = 400 \times \frac{2}{100} = 8 V$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359569 The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11} V .$ The electric field at the point is

1

4 π ε_{0} Q \times 10^{22} V / m

2

12 π ε_{0} Q \times 10^{22} V / m

3

4 π ε_{0} Q \times 10^{20} V / m

4

12 π ε_{0} Q \times 10^{20} V / m

Explanation:

As potential at a point due to a point charge is
$V = \frac{1}{4 π ε_{0}} \times \frac{Q}{r}$
The electric field is
$E = \frac{1}{4 π ε_{0}} \times \frac{Q}{r^{2}}$
Given that $V = Q \times 10^{11}$
From the above equations
$E = 4 π ε_{o} Q \times 10^{22} V m$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359570 The electric potential at a point $(x, y, z)$ is given by $V = - x^{2} y - x z^{3} + 4$ The electric field $\vec{E}$ at that point is

1

\vec{E} = \hat{i} (2 x y + z^{3}) + \hat{j} x^{2} + \hat{k} 3 x z^{2}

2

\vec{E} = \hat{i} (2 x y - z^{3}) + \hat{j} x y^{2} + \hat{k} 3 z^{2} x

3

\vec{E} = \hat{i} z^{3} + \hat{j} x y z + \hat{k} z^{2}

4

\vec{E} = \hat{i} 2 x y + \hat{j} (x^{2} + y^{2}) + \hat{k} (3 x z - y^{2})

Explanation:

$E_{x} = - \frac{d V}{d x} = - [- 2 x y - z^{3}]$
$E_{y} = - \frac{d V}{d y} = - [- x^{2}]$
$E_{z} = - \frac{d V}{d z} = - [- 3 x z^{2}]$
$∴ \vec{E} = E_{x} \hat{i} + E_{y} \hat{j} + E_{z} \hat{k} = (2 x y + z^{3}) \hat{i} + x^{2} \hat{j} + 3 x z^{2} \hat{k}$

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