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Electrostatic Potentials and Capacitance

268149 The equivalent capacitance $C_{A B}$ of the circuit shown in the figure is

1

\frac{5}{4} C

2

\frac{4}{5} C

3

2 C

C

4

C

Electrostatic Potentials and Capacitance

268075 The capacity of a parallel plate condenser consisting of two plates each $10 cm$ square and are seperated by a distance of $2 mm$ is (Take air as the medium between the plates)

1

8.85 \times 10^{- 13} F

2

4.42 \times 10^{- 12} F

3

44.25 \times 10^{- 12} F

4

88.5 \times 10^{- 13} F

Explanation:

$C = \frac{ε_{0} A}{d}$

Electrostatic Potentials and Capacitance

268076 Sixty four spherical drops each of radius $2 cm$ and carrying 5C charge combine to form a bigger drop. Its capacity is

1

\frac{8}{9} \times 10^{- 11} F

2

90 \times

3

1.1 \times 10^{- 11} F

4

9 \times 10^{11} F

Explanation:

$C^{1} = n^{1}$

Electrostatic Potentials and Capacitance

268077 A highly conducting sheet ofaluminium foil of negligible thickness is placed between the plates of a parallel plate capacitor. The foil is parallel to the plates. If the capacitance before the insertion of foil was $10 μ F$ , its value after the insertion of foil will be

1

20 μ F

2

10 μ F

3

5 μ F

4 Zero

Explanation:

$C$

Electrostatic Potentials and Capacitance

268078 Two metal platesare separated by a distance d in a parallel plate condenser. A metal plate of thickness $t$ and of the same area is inserted between the condenser plates. The value of capacitance increases by times

1

\frac{d - t}{d}

2

(1 - \frac{t}{d})

3

(t - \frac{t}{d})

4

\frac{1}{(1 - \frac{t}{d})}

Explanation:

$C = \frac{ε_{0} A}{d - t + \frac{t}{k}}; k = \infty$

Electrostatic Potentials and Capacitance

268149 The equivalent capacitance $C_{A B}$ of the circuit shown in the figure is

1

\frac{5}{4} C

2

\frac{4}{5} C

3

2 C

C

4

C

Electrostatic Potentials and Capacitance

268075 The capacity of a parallel plate condenser consisting of two plates each $10 cm$ square and are seperated by a distance of $2 mm$ is (Take air as the medium between the plates)

1

8.85 \times 10^{- 13} F

2

4.42 \times 10^{- 12} F

3

44.25 \times 10^{- 12} F

4

88.5 \times 10^{- 13} F

Explanation:

$C = \frac{ε_{0} A}{d}$

Electrostatic Potentials and Capacitance

268076 Sixty four spherical drops each of radius $2 cm$ and carrying 5C charge combine to form a bigger drop. Its capacity is

1

\frac{8}{9} \times 10^{- 11} F

2

90 \times

3

1.1 \times 10^{- 11} F

4

9 \times 10^{11} F

Explanation:

$C^{1} = n^{1}$

Electrostatic Potentials and Capacitance

268077 A highly conducting sheet ofaluminium foil of negligible thickness is placed between the plates of a parallel plate capacitor. The foil is parallel to the plates. If the capacitance before the insertion of foil was $10 μ F$ , its value after the insertion of foil will be

1

20 μ F

2

10 μ F

3

5 μ F

4 Zero

Explanation:

$C$

Electrostatic Potentials and Capacitance

268078 Two metal platesare separated by a distance d in a parallel plate condenser. A metal plate of thickness $t$ and of the same area is inserted between the condenser plates. The value of capacitance increases by times

1

\frac{d - t}{d}

2

(1 - \frac{t}{d})

3

(t - \frac{t}{d})

4

\frac{1}{(1 - \frac{t}{d})}

Explanation:

$C = \frac{ε_{0} A}{d - t + \frac{t}{k}}; k = \infty$

Electrostatic Potentials and Capacitance

268149 The equivalent capacitance $C_{A B}$ of the circuit shown in the figure is

1

\frac{5}{4} C

2

\frac{4}{5} C

3

2 C

C

4

C

Electrostatic Potentials and Capacitance

268075 The capacity of a parallel plate condenser consisting of two plates each $10 cm$ square and are seperated by a distance of $2 mm$ is (Take air as the medium between the plates)

1

8.85 \times 10^{- 13} F

2

4.42 \times 10^{- 12} F

3

44.25 \times 10^{- 12} F

4

88.5 \times 10^{- 13} F

Explanation:

$C = \frac{ε_{0} A}{d}$

Electrostatic Potentials and Capacitance

268076 Sixty four spherical drops each of radius $2 cm$ and carrying 5C charge combine to form a bigger drop. Its capacity is

1

\frac{8}{9} \times 10^{- 11} F

2

90 \times

3

1.1 \times 10^{- 11} F

4

9 \times 10^{11} F

Explanation:

$C^{1} = n^{1}$

Electrostatic Potentials and Capacitance

268077 A highly conducting sheet ofaluminium foil of negligible thickness is placed between the plates of a parallel plate capacitor. The foil is parallel to the plates. If the capacitance before the insertion of foil was $10 μ F$ , its value after the insertion of foil will be

1

20 μ F

2

10 μ F

3

5 μ F

4 Zero

Explanation:

$C$

Electrostatic Potentials and Capacitance

268078 Two metal platesare separated by a distance d in a parallel plate condenser. A metal plate of thickness $t$ and of the same area is inserted between the condenser plates. The value of capacitance increases by times

1

\frac{d - t}{d}

2

(1 - \frac{t}{d})

3

(t - \frac{t}{d})

4

\frac{1}{(1 - \frac{t}{d})}

Explanation:

$C = \frac{ε_{0} A}{d - t + \frac{t}{k}}; k = \infty$

Electrostatic Potentials and Capacitance

268149 The equivalent capacitance $C_{A B}$ of the circuit shown in the figure is

1

\frac{5}{4} C

2

\frac{4}{5} C

3

2 C

C

4

C

Electrostatic Potentials and Capacitance

268075 The capacity of a parallel plate condenser consisting of two plates each $10 cm$ square and are seperated by a distance of $2 mm$ is (Take air as the medium between the plates)

1

8.85 \times 10^{- 13} F

2

4.42 \times 10^{- 12} F

3

44.25 \times 10^{- 12} F

4

88.5 \times 10^{- 13} F

Explanation:

$C = \frac{ε_{0} A}{d}$

Electrostatic Potentials and Capacitance

268076 Sixty four spherical drops each of radius $2 cm$ and carrying 5C charge combine to form a bigger drop. Its capacity is

1

\frac{8}{9} \times 10^{- 11} F

2

90 \times

3

1.1 \times 10^{- 11} F

4

9 \times 10^{11} F

Explanation:

$C^{1} = n^{1}$

Electrostatic Potentials and Capacitance

268077 A highly conducting sheet ofaluminium foil of negligible thickness is placed between the plates of a parallel plate capacitor. The foil is parallel to the plates. If the capacitance before the insertion of foil was $10 μ F$ , its value after the insertion of foil will be

1

20 μ F

2

10 μ F

3

5 μ F

4 Zero

Explanation:

$C$

Electrostatic Potentials and Capacitance

268078 Two metal platesare separated by a distance d in a parallel plate condenser. A metal plate of thickness $t$ and of the same area is inserted between the condenser plates. The value of capacitance increases by times

1

\frac{d - t}{d}

2

(1 - \frac{t}{d})

3

(t - \frac{t}{d})

4

\frac{1}{(1 - \frac{t}{d})}

Explanation:

$C = \frac{ε_{0} A}{d - t + \frac{t}{k}}; k = \infty$

Electrostatic Potentials and Capacitance

268149 The equivalent capacitance $C_{A B}$ of the circuit shown in the figure is

1

\frac{5}{4} C

2

\frac{4}{5} C

3

2 C

C

4

C

Electrostatic Potentials and Capacitance

268075 The capacity of a parallel plate condenser consisting of two plates each $10 cm$ square and are seperated by a distance of $2 mm$ is (Take air as the medium between the plates)

1

8.85 \times 10^{- 13} F

2

4.42 \times 10^{- 12} F

3

44.25 \times 10^{- 12} F

4

88.5 \times 10^{- 13} F

Explanation:

$C = \frac{ε_{0} A}{d}$

Electrostatic Potentials and Capacitance

268076 Sixty four spherical drops each of radius $2 cm$ and carrying 5C charge combine to form a bigger drop. Its capacity is

1

\frac{8}{9} \times 10^{- 11} F

2

90 \times

3

1.1 \times 10^{- 11} F

4

9 \times 10^{11} F

Explanation:

$C^{1} = n^{1}$

Electrostatic Potentials and Capacitance

268077 A highly conducting sheet ofaluminium foil of negligible thickness is placed between the plates of a parallel plate capacitor. The foil is parallel to the plates. If the capacitance before the insertion of foil was $10 μ F$ , its value after the insertion of foil will be

1

20 μ F

2

10 μ F

3

5 μ F

4 Zero

Explanation:

$C$

Electrostatic Potentials and Capacitance

268078 Two metal platesare separated by a distance d in a parallel plate condenser. A metal plate of thickness $t$ and of the same area is inserted between the condenser plates. The value of capacitance increases by times

1

\frac{d - t}{d}

2

(1 - \frac{t}{d})

3

(t - \frac{t}{d})

4

\frac{1}{(1 - \frac{t}{d})}

Explanation:

$C = \frac{ε_{0} A}{d - t + \frac{t}{k}}; k = \infty$