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Electrostatic Potentials and Capacitance

272260 On decreasing the distance between the plates of a parallel plate capacitor, its capacitance

1 remains unaffected

2 decreases

3 first increases then decreases.

4 increases

Explanation:

(d) Since capacitance $C = \frac{ε_{0} A}{d}$ , as d decreases capacitance increases.

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272261 If in a parallel plate capacitor, which is connected to a battery, we fill dielectrics in whole space of its plates, then which of the following increases?

1

Q

and

V

2

V

and

E

3 E and C

4

Q

and

C

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272262 When air in a capacitor is replaced by a medium of dielectric constant $K$ , the capacity

1 decreases

K

times

2 increases

K^{2}

times

3 increases

K

times

4 remains constant

Explanation:

(b) $C_{medium} = K \times C_{air}$

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272263 A parallel plate capacitor with air between the plates has a capacitance of 8 pF . Calculate the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant. $(ε_{r} = 6)$

1 72 pF

2 81 pF

3 84 pF

4 96 pF

Explanation:

(d) Capacity of parallel plate capacitor
$C = \frac{ε_{r} ε_{0} A}{d} (For air ε_{r} = i)$
So, $\frac{ε_{a} A}{d} = 8 \times 10^{- 12}$
If $d \to \frac{d}{2}$ and $ε_{T} \to 6$ then new capacitance
$C^{'} = 6 \times \frac{ε_{0} A}{d / 2} = 12 \frac{ε_{0} A}{d} = 12 \times 8 pF = 96 pF$

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272260 On decreasing the distance between the plates of a parallel plate capacitor, its capacitance

1 remains unaffected

2 decreases

3 first increases then decreases.

4 increases

Explanation:

(d) Since capacitance $C = \frac{ε_{0} A}{d}$ , as d decreases capacitance increases.

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272261 If in a parallel plate capacitor, which is connected to a battery, we fill dielectrics in whole space of its plates, then which of the following increases?

1

Q

and

V

2

V

and

E

3 E and C

4

Q

and

C

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272262 When air in a capacitor is replaced by a medium of dielectric constant $K$ , the capacity

1 decreases

K

times

2 increases

K^{2}

times

3 increases

K

times

4 remains constant

Explanation:

(b) $C_{medium} = K \times C_{air}$

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272263 A parallel plate capacitor with air between the plates has a capacitance of 8 pF . Calculate the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant. $(ε_{r} = 6)$

1 72 pF

2 81 pF

3 84 pF

4 96 pF

Explanation:

(d) Capacity of parallel plate capacitor
$C = \frac{ε_{r} ε_{0} A}{d} (For air ε_{r} = i)$
So, $\frac{ε_{a} A}{d} = 8 \times 10^{- 12}$
If $d \to \frac{d}{2}$ and $ε_{T} \to 6$ then new capacitance
$C^{'} = 6 \times \frac{ε_{0} A}{d / 2} = 12 \frac{ε_{0} A}{d} = 12 \times 8 pF = 96 pF$

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272260 On decreasing the distance between the plates of a parallel plate capacitor, its capacitance

1 remains unaffected

2 decreases

3 first increases then decreases.

4 increases

Explanation:

(d) Since capacitance $C = \frac{ε_{0} A}{d}$ , as d decreases capacitance increases.

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272261 If in a parallel plate capacitor, which is connected to a battery, we fill dielectrics in whole space of its plates, then which of the following increases?

1

Q

and

V

2

V

and

E

3 E and C

4

Q

and

C

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272262 When air in a capacitor is replaced by a medium of dielectric constant $K$ , the capacity

1 decreases

K

times

2 increases

K^{2}

times

3 increases

K

times

4 remains constant

Explanation:

(b) $C_{medium} = K \times C_{air}$

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272263 A parallel plate capacitor with air between the plates has a capacitance of 8 pF . Calculate the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant. $(ε_{r} = 6)$

1 72 pF

2 81 pF

3 84 pF

4 96 pF

Explanation:

(d) Capacity of parallel plate capacitor
$C = \frac{ε_{r} ε_{0} A}{d} (For air ε_{r} = i)$
So, $\frac{ε_{a} A}{d} = 8 \times 10^{- 12}$
If $d \to \frac{d}{2}$ and $ε_{T} \to 6$ then new capacitance
$C^{'} = 6 \times \frac{ε_{0} A}{d / 2} = 12 \frac{ε_{0} A}{d} = 12 \times 8 pF = 96 pF$

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272260 On decreasing the distance between the plates of a parallel plate capacitor, its capacitance

1 remains unaffected

2 decreases

3 first increases then decreases.

4 increases

Explanation:

(d) Since capacitance $C = \frac{ε_{0} A}{d}$ , as d decreases capacitance increases.

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272261 If in a parallel plate capacitor, which is connected to a battery, we fill dielectrics in whole space of its plates, then which of the following increases?

1

Q

and

V

2

V

and

E

3 E and C

4

Q

and

C

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272262 When air in a capacitor is replaced by a medium of dielectric constant $K$ , the capacity

1 decreases

K

times

2 increases

K^{2}

times

3 increases

K

times

4 remains constant

Explanation:

(b) $C_{medium} = K \times C_{air}$

NCERT Page-75/N-69

Electrostatic Potentials and Capacitance

272263 A parallel plate capacitor with air between the plates has a capacitance of 8 pF . Calculate the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant. $(ε_{r} = 6)$

1 72 pF

2 81 pF

3 84 pF

4 96 pF

Explanation:

(d) Capacity of parallel plate capacitor
$C = \frac{ε_{r} ε_{0} A}{d} (For air ε_{r} = i)$
So, $\frac{ε_{a} A}{d} = 8 \times 10^{- 12}$
If $d \to \frac{d}{2}$ and $ε_{T} \to 6$ then new capacitance
$C^{'} = 6 \times \frac{ε_{0} A}{d / 2} = 12 \frac{ε_{0} A}{d} = 12 \times 8 pF = 96 pF$

NCERT Page-75/N-69

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