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Electrostatic Potentials and Capacitance

268115 An infinite number of identical capacitorseach of capacitance $1 mF$ are connected as shown in the figure. Then the equivalent capacitance between $A$ and $B$ is

1

1 mF

2

2 mF

3

1 / 2 mF

4

0.75 mF

Explanation:

$C_{R} = C + \frac{C}{2} + \frac{C}{4} + \dots .$ .

Electrostatic Potentials and Capacitance

268101 The resultant capacity between the points $P$ and $Q$ of the given figure is

1

4 μ F

2

\frac{16}{3} μ F

3

1.6 μ F

4

1 μ F

Explanation:

$C_{1} = \frac{4 \times 4}{4 + 4}; C_{2} = 2 μ F; C_{eff} = C_{1} + C_{2}$

Electrostatic Potentials and Capacitance

268125 'A' and ' $B$ ' are two condensers of capacities $2 μ F$ and $4 μ F$ . They arecharged to potential differences of $12 V$ and $6 V$ respectively. If they are now connected (+ve to +ve), the charge that flows through the connecting wire is

1

24 μ C

fromA to

B

2

8 μ C

fromA to

B

3

8 μ C

fromB to

A

4

24 μ C

fromB to

A

Explanation:

Chargeflow $Δ Q = C_{1} [V_{1} - V_{0}]$
where $V_{o} \to$ common potential
$t \to$ thickness of metal sheet

Electrostatic Potentials and Capacitance

268126 Force of attraction between the plates of a parallel plate capacitor is

1

\frac{q^{2}}{2 ε_{0} A}

2

\frac{q^{2}}{ε_{0} A}

3

\frac{q}{2 ε_{0} A}

4

\frac{q^{2}}{2 ε_{0} A^{2}}

Explanation:

$q = C V = \frac{C_{1} C_{2}}{C_{1} + C_{2}} . V . = V_{o} = \frac{q}{C_{1}}$

Electrostatic Potentials and Capacitance

268115 An infinite number of identical capacitorseach of capacitance $1 mF$ are connected as shown in the figure. Then the equivalent capacitance between $A$ and $B$ is

1

1 mF

2

2 mF

3

1 / 2 mF

4

0.75 mF

Explanation:

$C_{R} = C + \frac{C}{2} + \frac{C}{4} + \dots .$ .

Electrostatic Potentials and Capacitance

268101 The resultant capacity between the points $P$ and $Q$ of the given figure is

1

4 μ F

2

\frac{16}{3} μ F

3

1.6 μ F

4

1 μ F

Explanation:

$C_{1} = \frac{4 \times 4}{4 + 4}; C_{2} = 2 μ F; C_{eff} = C_{1} + C_{2}$

Electrostatic Potentials and Capacitance

268125 'A' and ' $B$ ' are two condensers of capacities $2 μ F$ and $4 μ F$ . They arecharged to potential differences of $12 V$ and $6 V$ respectively. If they are now connected (+ve to +ve), the charge that flows through the connecting wire is

1

24 μ C

fromA to

B

2

8 μ C

fromA to

B

3

8 μ C

fromB to

A

4

24 μ C

fromB to

A

Explanation:

Chargeflow $Δ Q = C_{1} [V_{1} - V_{0}]$
where $V_{o} \to$ common potential
$t \to$ thickness of metal sheet

Electrostatic Potentials and Capacitance

268126 Force of attraction between the plates of a parallel plate capacitor is

1

\frac{q^{2}}{2 ε_{0} A}

2

\frac{q^{2}}{ε_{0} A}

3

\frac{q}{2 ε_{0} A}

4

\frac{q^{2}}{2 ε_{0} A^{2}}

Explanation:

$q = C V = \frac{C_{1} C_{2}}{C_{1} + C_{2}} . V . = V_{o} = \frac{q}{C_{1}}$

Electrostatic Potentials and Capacitance

268115 An infinite number of identical capacitorseach of capacitance $1 mF$ are connected as shown in the figure. Then the equivalent capacitance between $A$ and $B$ is

1

1 mF

2

2 mF

3

1 / 2 mF

4

0.75 mF

Explanation:

$C_{R} = C + \frac{C}{2} + \frac{C}{4} + \dots .$ .

Electrostatic Potentials and Capacitance

268101 The resultant capacity between the points $P$ and $Q$ of the given figure is

1

4 μ F

2

\frac{16}{3} μ F

3

1.6 μ F

4

1 μ F

Explanation:

$C_{1} = \frac{4 \times 4}{4 + 4}; C_{2} = 2 μ F; C_{eff} = C_{1} + C_{2}$

Electrostatic Potentials and Capacitance

268125 'A' and ' $B$ ' are two condensers of capacities $2 μ F$ and $4 μ F$ . They arecharged to potential differences of $12 V$ and $6 V$ respectively. If they are now connected (+ve to +ve), the charge that flows through the connecting wire is

1

24 μ C

fromA to

B

2

8 μ C

fromA to

B

3

8 μ C

fromB to

A

4

24 μ C

fromB to

A

Explanation:

Chargeflow $Δ Q = C_{1} [V_{1} - V_{0}]$
where $V_{o} \to$ common potential
$t \to$ thickness of metal sheet

Electrostatic Potentials and Capacitance

268126 Force of attraction between the plates of a parallel plate capacitor is

1

\frac{q^{2}}{2 ε_{0} A}

2

\frac{q^{2}}{ε_{0} A}

3

\frac{q}{2 ε_{0} A}

4

\frac{q^{2}}{2 ε_{0} A^{2}}

Explanation:

$q = C V = \frac{C_{1} C_{2}}{C_{1} + C_{2}} . V . = V_{o} = \frac{q}{C_{1}}$

Electrostatic Potentials and Capacitance

268115 An infinite number of identical capacitorseach of capacitance $1 mF$ are connected as shown in the figure. Then the equivalent capacitance between $A$ and $B$ is

1

1 mF

2

2 mF

3

1 / 2 mF

4

0.75 mF

Explanation:

$C_{R} = C + \frac{C}{2} + \frac{C}{4} + \dots .$ .

Electrostatic Potentials and Capacitance

268101 The resultant capacity between the points $P$ and $Q$ of the given figure is

1

4 μ F

2

\frac{16}{3} μ F

3

1.6 μ F

4

1 μ F

Explanation:

$C_{1} = \frac{4 \times 4}{4 + 4}; C_{2} = 2 μ F; C_{eff} = C_{1} + C_{2}$

Electrostatic Potentials and Capacitance

268125 'A' and ' $B$ ' are two condensers of capacities $2 μ F$ and $4 μ F$ . They arecharged to potential differences of $12 V$ and $6 V$ respectively. If they are now connected (+ve to +ve), the charge that flows through the connecting wire is

1

24 μ C

fromA to

B

2

8 μ C

fromA to

B

3

8 μ C

fromB to

A

4

24 μ C

fromB to

A

Explanation:

Chargeflow $Δ Q = C_{1} [V_{1} - V_{0}]$
where $V_{o} \to$ common potential
$t \to$ thickness of metal sheet

Electrostatic Potentials and Capacitance

268126 Force of attraction between the plates of a parallel plate capacitor is

1

\frac{q^{2}}{2 ε_{0} A}

2

\frac{q^{2}}{ε_{0} A}

3

\frac{q}{2 ε_{0} A}

4

\frac{q^{2}}{2 ε_{0} A^{2}}

Explanation:

$q = C V = \frac{C_{1} C_{2}}{C_{1} + C_{2}} . V . = V_{o} = \frac{q}{C_{1}}$

Question Bank Series

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Question Bank Series

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