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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359220 A parallel plate air capacitor has a capacitance $18 μ F$ . If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes $54 μ F$ . The dielectric constant of the medium is

1

4

2

9

3

12

4

2

Explanation:

Let $C_{0}$ be the initial capacitance.
$C_{0} = \frac{ε_{0} A}{d} = 18 (1)$
Let $K$ be the dielectric constant of the medium
$C = \frac{K ε_{0} A}{3 d} = 54 (2)$
On dividing Eq. (2) by Eq. (1), we get
$\frac{K}{3} = \frac{54}{18}$
$∴$ Dielectric constant, $K = 9$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359221 A parallel plate capacitor has capacitance $C$ . If it is equally filled with parallel layers of materials of dielectric constant $K_{1}$ and $K_{2}$ its capacity becomes $C_{1}$ . The ratio of $C_{1}$ and $C$ is

1

K_{1} + K_{2}

2

\frac{K_{1} K_{2}}{K_{1} + K_{2}}

3

\frac{K_{1} + K_{2}}{K_{1} K_{2}}

4

\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}

Explanation:

The capacitance of a parallel plate capacitor is
$C = \frac{ε_{0} A}{d}$
where $A$ is the area of each plate and $d$ is the separation between the plates.
When it is equally filled with layers of materials of dielectric constant $K_{1}$ and $K_{2}$ as shown in figure, then the arrangement is equivalent to two capacitors each of area $A$ and separation $d / 2$ connected in series.
supporting img
$∴ \frac{1}{C_{1}} = \frac{d / 2}{ε_{0} K_{1} A} + \frac{d / 2}{ε_{0} K_{2} A}$
or $C_{1} = \frac{ε_{0} A}{d} [\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}] ∴ \frac{C_{1}}{C} = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359222 An air capacitor of capacity $C = 20 μ F$ is connected to a constant voltage battery of 6 $V$ . Now the space between the plates is filled with a liquid of dielectric constant 2. The charge that flows now from battery to the capacitor is

1

120 μ C

2

240 μ C

3

180 μ C

4

80 μ C

Explanation:

Intially charge on the capacitor
$Q = 20 \times 6 = 60 μ C$
Finally chagre on the capacitor
$Q^{'} = (2 \times 20) \times 6 = 240 μ C$
So charge supplied by the battery after placing dielectric is
$= Q^{'} - Q = 180 μ C$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359223 A voltmeter reads 4 $V$ when connected parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between plates for the same configuration, voltmeter reads 2 $V$ . What is the dielectric constant of the material?

1

8

2

0.5

3

10

4

2

Explanation:

Let $V_{0}$ be the potential difference between the plates of an air filled parallel plate capacitor. When a dielectric slab is introduced between its plates, its potential difference will become
$V = \frac{V_{0}}{K}$
where K is the dielectric constant of the slab.
$∴ K = \frac{V_{0}}{V}$
Here $V_{0} = 4 V, V = 2 V$
$∴ K = \frac{4 V}{2 V} = 2$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359220 A parallel plate air capacitor has a capacitance $18 μ F$ . If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes $54 μ F$ . The dielectric constant of the medium is

1

4

2

9

3

12

4

2

Explanation:

Let $C_{0}$ be the initial capacitance.
$C_{0} = \frac{ε_{0} A}{d} = 18 (1)$
Let $K$ be the dielectric constant of the medium
$C = \frac{K ε_{0} A}{3 d} = 54 (2)$
On dividing Eq. (2) by Eq. (1), we get
$\frac{K}{3} = \frac{54}{18}$
$∴$ Dielectric constant, $K = 9$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359221 A parallel plate capacitor has capacitance $C$ . If it is equally filled with parallel layers of materials of dielectric constant $K_{1}$ and $K_{2}$ its capacity becomes $C_{1}$ . The ratio of $C_{1}$ and $C$ is

1

K_{1} + K_{2}

2

\frac{K_{1} K_{2}}{K_{1} + K_{2}}

3

\frac{K_{1} + K_{2}}{K_{1} K_{2}}

4

\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}

Explanation:

The capacitance of a parallel plate capacitor is
$C = \frac{ε_{0} A}{d}$
where $A$ is the area of each plate and $d$ is the separation between the plates.
When it is equally filled with layers of materials of dielectric constant $K_{1}$ and $K_{2}$ as shown in figure, then the arrangement is equivalent to two capacitors each of area $A$ and separation $d / 2$ connected in series.
supporting img
$∴ \frac{1}{C_{1}} = \frac{d / 2}{ε_{0} K_{1} A} + \frac{d / 2}{ε_{0} K_{2} A}$
or $C_{1} = \frac{ε_{0} A}{d} [\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}] ∴ \frac{C_{1}}{C} = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359222 An air capacitor of capacity $C = 20 μ F$ is connected to a constant voltage battery of 6 $V$ . Now the space between the plates is filled with a liquid of dielectric constant 2. The charge that flows now from battery to the capacitor is

1

120 μ C

2

240 μ C

3

180 μ C

4

80 μ C

Explanation:

Intially charge on the capacitor
$Q = 20 \times 6 = 60 μ C$
Finally chagre on the capacitor
$Q^{'} = (2 \times 20) \times 6 = 240 μ C$
So charge supplied by the battery after placing dielectric is
$= Q^{'} - Q = 180 μ C$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359223 A voltmeter reads 4 $V$ when connected parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between plates for the same configuration, voltmeter reads 2 $V$ . What is the dielectric constant of the material?

1

8

2

0.5

3

10

4

2

Explanation:

Let $V_{0}$ be the potential difference between the plates of an air filled parallel plate capacitor. When a dielectric slab is introduced between its plates, its potential difference will become
$V = \frac{V_{0}}{K}$
where K is the dielectric constant of the slab.
$∴ K = \frac{V_{0}}{V}$
Here $V_{0} = 4 V, V = 2 V$
$∴ K = \frac{4 V}{2 V} = 2$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359220 A parallel plate air capacitor has a capacitance $18 μ F$ . If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes $54 μ F$ . The dielectric constant of the medium is

1

4

2

9

3

12

4

2

Explanation:

Let $C_{0}$ be the initial capacitance.
$C_{0} = \frac{ε_{0} A}{d} = 18 (1)$
Let $K$ be the dielectric constant of the medium
$C = \frac{K ε_{0} A}{3 d} = 54 (2)$
On dividing Eq. (2) by Eq. (1), we get
$\frac{K}{3} = \frac{54}{18}$
$∴$ Dielectric constant, $K = 9$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359221 A parallel plate capacitor has capacitance $C$ . If it is equally filled with parallel layers of materials of dielectric constant $K_{1}$ and $K_{2}$ its capacity becomes $C_{1}$ . The ratio of $C_{1}$ and $C$ is

1

K_{1} + K_{2}

2

\frac{K_{1} K_{2}}{K_{1} + K_{2}}

3

\frac{K_{1} + K_{2}}{K_{1} K_{2}}

4

\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}

Explanation:

The capacitance of a parallel plate capacitor is
$C = \frac{ε_{0} A}{d}$
where $A$ is the area of each plate and $d$ is the separation between the plates.
When it is equally filled with layers of materials of dielectric constant $K_{1}$ and $K_{2}$ as shown in figure, then the arrangement is equivalent to two capacitors each of area $A$ and separation $d / 2$ connected in series.
supporting img
$∴ \frac{1}{C_{1}} = \frac{d / 2}{ε_{0} K_{1} A} + \frac{d / 2}{ε_{0} K_{2} A}$
or $C_{1} = \frac{ε_{0} A}{d} [\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}] ∴ \frac{C_{1}}{C} = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359222 An air capacitor of capacity $C = 20 μ F$ is connected to a constant voltage battery of 6 $V$ . Now the space between the plates is filled with a liquid of dielectric constant 2. The charge that flows now from battery to the capacitor is

1

120 μ C

2

240 μ C

3

180 μ C

4

80 μ C

Explanation:

Intially charge on the capacitor
$Q = 20 \times 6 = 60 μ C$
Finally chagre on the capacitor
$Q^{'} = (2 \times 20) \times 6 = 240 μ C$
So charge supplied by the battery after placing dielectric is
$= Q^{'} - Q = 180 μ C$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359223 A voltmeter reads 4 $V$ when connected parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between plates for the same configuration, voltmeter reads 2 $V$ . What is the dielectric constant of the material?

1

8

2

0.5

3

10

4

2

Explanation:

Let $V_{0}$ be the potential difference between the plates of an air filled parallel plate capacitor. When a dielectric slab is introduced between its plates, its potential difference will become
$V = \frac{V_{0}}{K}$
where K is the dielectric constant of the slab.
$∴ K = \frac{V_{0}}{V}$
Here $V_{0} = 4 V, V = 2 V$
$∴ K = \frac{4 V}{2 V} = 2$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359220 A parallel plate air capacitor has a capacitance $18 μ F$ . If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes $54 μ F$ . The dielectric constant of the medium is

1

4

2

9

3

12

4

2

Explanation:

Let $C_{0}$ be the initial capacitance.
$C_{0} = \frac{ε_{0} A}{d} = 18 (1)$
Let $K$ be the dielectric constant of the medium
$C = \frac{K ε_{0} A}{3 d} = 54 (2)$
On dividing Eq. (2) by Eq. (1), we get
$\frac{K}{3} = \frac{54}{18}$
$∴$ Dielectric constant, $K = 9$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359221 A parallel plate capacitor has capacitance $C$ . If it is equally filled with parallel layers of materials of dielectric constant $K_{1}$ and $K_{2}$ its capacity becomes $C_{1}$ . The ratio of $C_{1}$ and $C$ is

1

K_{1} + K_{2}

2

\frac{K_{1} K_{2}}{K_{1} + K_{2}}

3

\frac{K_{1} + K_{2}}{K_{1} K_{2}}

4

\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}

Explanation:

The capacitance of a parallel plate capacitor is
$C = \frac{ε_{0} A}{d}$
where $A$ is the area of each plate and $d$ is the separation between the plates.
When it is equally filled with layers of materials of dielectric constant $K_{1}$ and $K_{2}$ as shown in figure, then the arrangement is equivalent to two capacitors each of area $A$ and separation $d / 2$ connected in series.
supporting img
$∴ \frac{1}{C_{1}} = \frac{d / 2}{ε_{0} K_{1} A} + \frac{d / 2}{ε_{0} K_{2} A}$
or $C_{1} = \frac{ε_{0} A}{d} [\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}] ∴ \frac{C_{1}}{C} = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359222 An air capacitor of capacity $C = 20 μ F$ is connected to a constant voltage battery of 6 $V$ . Now the space between the plates is filled with a liquid of dielectric constant 2. The charge that flows now from battery to the capacitor is

1

120 μ C

2

240 μ C

3

180 μ C

4

80 μ C

Explanation:

Intially charge on the capacitor
$Q = 20 \times 6 = 60 μ C$
Finally chagre on the capacitor
$Q^{'} = (2 \times 20) \times 6 = 240 μ C$
So charge supplied by the battery after placing dielectric is
$= Q^{'} - Q = 180 μ C$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359223 A voltmeter reads 4 $V$ when connected parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between plates for the same configuration, voltmeter reads 2 $V$ . What is the dielectric constant of the material?

1

8

2

0.5

3

10

4

2

Explanation:

Let $V_{0}$ be the potential difference between the plates of an air filled parallel plate capacitor. When a dielectric slab is introduced between its plates, its potential difference will become
$V = \frac{V_{0}}{K}$
where K is the dielectric constant of the slab.
$∴ K = \frac{V_{0}}{V}$
Here $V_{0} = 4 V, V = 2 V$
$∴ K = \frac{4 V}{2 V} = 2$

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