QBManager

Capacitance

165986 A parallel plate capacitor of area $A$ , plate separation $d$ and capacitance $C$ is filled with three dielectric materials having dielectric constants $k_{1}, k_{2}$ and $k_{3}$ as shown. If a single dielectric material is to be used to have the same capacitance $C$ in the capacitor, then its dielectric constant $k$ is given by

1

K = K_{1} + K_{2} + 2 K_{3}

2

K = \frac{K_{1} K_{2}}{K_{1} + K_{2}} + 2 K_{3}

3

\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{2 K_{3}}

4

\frac{1}{K} = \frac{1}{K_{1} + K_{2}} + \frac{1}{2 K_{3}}

Explanation:

: In the given figure, we can divide it into three different capacitors. Let the capacitance with dielectric constant $K_{1}, K_{2}$ and $K_{3}$ be $C_{1}, C_{2}$ and $C_{3}$ capacity, the width of capacitor $= \frac{d}{2}$ and area is $A / 2$
We have, $C_{1} = \frac{ε_{0} (A / 2) K_{1}}{(d / 2)} = \frac{ε_{0} {AK}_{1}}{d}$
$C_{2} = \frac{ε_{0} (A / 2) K_{2}}{(d / 2)} = \frac{ε_{0} {AK}_{2}}{d}$
$C_{3} = \frac{ε_{0} {AK}_{3}}{(d / 2)} = \frac{2 ε_{0} {AK}_{3}}{d}$
The capacitors $C_{1}$ and $C_{2}$ are in parallel
$∴ C^{'} = C_{1} + C_{2} = \frac{ε_{0} A}{d} (K_{1} + K_{2})$
This combination is in series with $C_{3}$
The net capacitance
$\frac{1}{C^{''}} = \frac{1}{C^{'}} + \frac{1}{C_{3}}$
$\frac{1}{C^{''}} = \frac{1}{\frac{ε_{0} A}{d} (K_{1} + K_{2})} + \frac{1}{\frac{2 ε_{0} {AK}_{3}}{d}}$
$\frac{1}{C^{''}} = \frac{d}{ε_{0} A (K_{1} + K_{2})} + \frac{d}{2 ε_{0} {AK}_{3}}$
$\frac{1}{C^{''}} = \frac{d}{ε_{0} A} [\frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}]$
Where, $(C^{''} = \frac{ε_{0} AK}{d})$
Then, $\frac{1}{K} = \frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}$

[UPSEE - 2015]

Capacitance

165987 A spherical condenser has inner and outer spheres of radii a and $b$ respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be

1 Zero

2

4 π ε_{0} a

3

4 π ε_{0} b

4

4 π ε_{0} a (\frac{b}{b - a})

Explanation:

: Given, the Radius of inner sphere $= a$ ,
The Radius of outer Sphere $= b$
The expression to calculate the capacitance when outer sphere is earthed.
$C_{1} = 4 π ε_{0} \frac{ab}{b - a}$
The expression to calculate the capacitance when inner sphere is earthed
$C_{2} = 4 π ε_{0} b + \frac{4 π ε_{0} ab}{b - a}$
$C_{2} = 4 π ε_{0} (\frac{b^{2}}{b - a})$
Difference between equation (i) and (ii)
$C_{2} - C_{1} = 4 π ε_{0} (\frac{b^{2}}{b - a}) - 4 π ε_{0} \frac{ab}{b - a}$
$C_{2} - C_{1} = 4 π ε_{0} b [\frac{b - a}{b - a}]$
$C_{2} - C_{1} = 4 π ε_{0} b$

[UPSEE - 2014]

Capacitance

165988 Capacitance of a capacitor made by a thin metal foil is $2 μ F$ . If the foils is folded with paper of thickness $0.15 mm$ , dielectric constant of paper is 2.5 and width of paper is $400 mm$ , the length of the foil will be

1

0.34 m

2

1.33 m

3

13.4 m

4

33.9 m

Explanation:

: Given, $C = 2 \times 10^{- 6} F$
Width of paper $(b) = 400 mm = 0.4 m$
Thickness of paper $(t) = 0.15 mm = 15 \times 10^{- 5} m$
Dielectric constant of paper $(K) = 2.5$
As capacitance for parallel plate capacitor is,
$C = \frac{K ε_{0} A}{d}$
$C = \frac{K ε_{0} (l \times b)}{t}$
$l = \frac{Ct}{K ε_{0} b}$
$l = \frac{2 \times 10^{- 6} \times 15 \times 10^{- 5}}{(2.5 \times 8.85 \times 10^{- 12} \times 0.4)} = 33.89 ⊔ 33.9$
$l = 33.9 m .$

[UPSEE - 2012]

Capacitance

165990 In the condenser shown in the circuit is charged to $5 V$ and left in the circuit, in $12 s$ the charge on the condenser will become :

1

\frac{10}{e} C

2

\frac{e}{10} C

3

\frac{10}{e^{2}} C

4

\frac{e^{2}}{10} C

5

= 2.718

Explanation:

: Given, $C = 2 F, V = 5 V, t = 12 \sec, R = 6 Ω$ Exponential decay of charge take place.
In $C - R$ Circuit, discharging take place.
$q = Q e^{- t / RC}$
Where, $R$ in resistance, $C$ is Capacitance and
$Q = CV$
Hence, $q = CV e^{- t / RC}$
$q = (2 \times 5 \times e^{- 12 / 6 \times 2}) Coulomb$
$q = \frac{10}{e} C$

[UPSEE - 2006

Capacitance

165991 A capacitor having capacitance $1 μ F$ with air is filled with two dielectrics as shown. How many times capacitance will increase?

1 12

2 6

3

8 / 3

4 3

Explanation:

: Given that,
Dielectric constant, $K_{1} = 8, K_{2} = 4$
We know that,
$C = \frac{ε_{0} A}{d}$
$∴ C_{1} = \frac{K_{1} ε_{0} A / 2}{d} = \frac{8}{2} \frac{ε_{0} A}{d} = 4 C$
$C_{2} = \frac{K_{2} ε_{0} A / 2}{d} = \frac{4}{2} \frac{ε_{0} A}{d} = 2 C$
Capacitance will increase due to parallel combination
$C^{'} = C_{1} + C_{2}$
$= 4 C + 2 C$
$= 6 C$

CGPET-2010]

Capacitance

165986 A parallel plate capacitor of area $A$ , plate separation $d$ and capacitance $C$ is filled with three dielectric materials having dielectric constants $k_{1}, k_{2}$ and $k_{3}$ as shown. If a single dielectric material is to be used to have the same capacitance $C$ in the capacitor, then its dielectric constant $k$ is given by

1

K = K_{1} + K_{2} + 2 K_{3}

2

K = \frac{K_{1} K_{2}}{K_{1} + K_{2}} + 2 K_{3}

3

\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{2 K_{3}}

4

\frac{1}{K} = \frac{1}{K_{1} + K_{2}} + \frac{1}{2 K_{3}}

Explanation:

: In the given figure, we can divide it into three different capacitors. Let the capacitance with dielectric constant $K_{1}, K_{2}$ and $K_{3}$ be $C_{1}, C_{2}$ and $C_{3}$ capacity, the width of capacitor $= \frac{d}{2}$ and area is $A / 2$
We have, $C_{1} = \frac{ε_{0} (A / 2) K_{1}}{(d / 2)} = \frac{ε_{0} {AK}_{1}}{d}$
$C_{2} = \frac{ε_{0} (A / 2) K_{2}}{(d / 2)} = \frac{ε_{0} {AK}_{2}}{d}$
$C_{3} = \frac{ε_{0} {AK}_{3}}{(d / 2)} = \frac{2 ε_{0} {AK}_{3}}{d}$
The capacitors $C_{1}$ and $C_{2}$ are in parallel
$∴ C^{'} = C_{1} + C_{2} = \frac{ε_{0} A}{d} (K_{1} + K_{2})$
This combination is in series with $C_{3}$
The net capacitance
$\frac{1}{C^{''}} = \frac{1}{C^{'}} + \frac{1}{C_{3}}$
$\frac{1}{C^{''}} = \frac{1}{\frac{ε_{0} A}{d} (K_{1} + K_{2})} + \frac{1}{\frac{2 ε_{0} {AK}_{3}}{d}}$
$\frac{1}{C^{''}} = \frac{d}{ε_{0} A (K_{1} + K_{2})} + \frac{d}{2 ε_{0} {AK}_{3}}$
$\frac{1}{C^{''}} = \frac{d}{ε_{0} A} [\frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}]$
Where, $(C^{''} = \frac{ε_{0} AK}{d})$
Then, $\frac{1}{K} = \frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}$

[UPSEE - 2015]

Capacitance

165987 A spherical condenser has inner and outer spheres of radii a and $b$ respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be

1 Zero

2

4 π ε_{0} a

3

4 π ε_{0} b

4

4 π ε_{0} a (\frac{b}{b - a})

Explanation:

: Given, the Radius of inner sphere $= a$ ,
The Radius of outer Sphere $= b$
The expression to calculate the capacitance when outer sphere is earthed.
$C_{1} = 4 π ε_{0} \frac{ab}{b - a}$
The expression to calculate the capacitance when inner sphere is earthed
$C_{2} = 4 π ε_{0} b + \frac{4 π ε_{0} ab}{b - a}$
$C_{2} = 4 π ε_{0} (\frac{b^{2}}{b - a})$
Difference between equation (i) and (ii)
$C_{2} - C_{1} = 4 π ε_{0} (\frac{b^{2}}{b - a}) - 4 π ε_{0} \frac{ab}{b - a}$
$C_{2} - C_{1} = 4 π ε_{0} b [\frac{b - a}{b - a}]$
$C_{2} - C_{1} = 4 π ε_{0} b$

[UPSEE - 2014]

Capacitance

165988 Capacitance of a capacitor made by a thin metal foil is $2 μ F$ . If the foils is folded with paper of thickness $0.15 mm$ , dielectric constant of paper is 2.5 and width of paper is $400 mm$ , the length of the foil will be

1

0.34 m

2

1.33 m

3

13.4 m

4

33.9 m

Explanation:

: Given, $C = 2 \times 10^{- 6} F$
Width of paper $(b) = 400 mm = 0.4 m$
Thickness of paper $(t) = 0.15 mm = 15 \times 10^{- 5} m$
Dielectric constant of paper $(K) = 2.5$
As capacitance for parallel plate capacitor is,
$C = \frac{K ε_{0} A}{d}$
$C = \frac{K ε_{0} (l \times b)}{t}$
$l = \frac{Ct}{K ε_{0} b}$
$l = \frac{2 \times 10^{- 6} \times 15 \times 10^{- 5}}{(2.5 \times 8.85 \times 10^{- 12} \times 0.4)} = 33.89 ⊔ 33.9$
$l = 33.9 m .$

[UPSEE - 2012]

Capacitance

165990 In the condenser shown in the circuit is charged to $5 V$ and left in the circuit, in $12 s$ the charge on the condenser will become :

1

\frac{10}{e} C

2

\frac{e}{10} C

3

\frac{10}{e^{2}} C

4

\frac{e^{2}}{10} C

5

= 2.718

Explanation:

: Given, $C = 2 F, V = 5 V, t = 12 \sec, R = 6 Ω$ Exponential decay of charge take place.
In $C - R$ Circuit, discharging take place.
$q = Q e^{- t / RC}$
Where, $R$ in resistance, $C$ is Capacitance and
$Q = CV$
Hence, $q = CV e^{- t / RC}$
$q = (2 \times 5 \times e^{- 12 / 6 \times 2}) Coulomb$
$q = \frac{10}{e} C$

[UPSEE - 2006

Capacitance

165991 A capacitor having capacitance $1 μ F$ with air is filled with two dielectrics as shown. How many times capacitance will increase?

1 12

2 6

3

8 / 3

4 3

Explanation:

: Given that,
Dielectric constant, $K_{1} = 8, K_{2} = 4$
We know that,
$C = \frac{ε_{0} A}{d}$
$∴ C_{1} = \frac{K_{1} ε_{0} A / 2}{d} = \frac{8}{2} \frac{ε_{0} A}{d} = 4 C$
$C_{2} = \frac{K_{2} ε_{0} A / 2}{d} = \frac{4}{2} \frac{ε_{0} A}{d} = 2 C$
Capacitance will increase due to parallel combination
$C^{'} = C_{1} + C_{2}$
$= 4 C + 2 C$
$= 6 C$

CGPET-2010]

Capacitance

165986 A parallel plate capacitor of area $A$ , plate separation $d$ and capacitance $C$ is filled with three dielectric materials having dielectric constants $k_{1}, k_{2}$ and $k_{3}$ as shown. If a single dielectric material is to be used to have the same capacitance $C$ in the capacitor, then its dielectric constant $k$ is given by

1

K = K_{1} + K_{2} + 2 K_{3}

2

K = \frac{K_{1} K_{2}}{K_{1} + K_{2}} + 2 K_{3}

3

\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{2 K_{3}}

4

\frac{1}{K} = \frac{1}{K_{1} + K_{2}} + \frac{1}{2 K_{3}}

Explanation:

: In the given figure, we can divide it into three different capacitors. Let the capacitance with dielectric constant $K_{1}, K_{2}$ and $K_{3}$ be $C_{1}, C_{2}$ and $C_{3}$ capacity, the width of capacitor $= \frac{d}{2}$ and area is $A / 2$
We have, $C_{1} = \frac{ε_{0} (A / 2) K_{1}}{(d / 2)} = \frac{ε_{0} {AK}_{1}}{d}$
$C_{2} = \frac{ε_{0} (A / 2) K_{2}}{(d / 2)} = \frac{ε_{0} {AK}_{2}}{d}$
$C_{3} = \frac{ε_{0} {AK}_{3}}{(d / 2)} = \frac{2 ε_{0} {AK}_{3}}{d}$
The capacitors $C_{1}$ and $C_{2}$ are in parallel
$∴ C^{'} = C_{1} + C_{2} = \frac{ε_{0} A}{d} (K_{1} + K_{2})$
This combination is in series with $C_{3}$
The net capacitance
$\frac{1}{C^{''}} = \frac{1}{C^{'}} + \frac{1}{C_{3}}$
$\frac{1}{C^{''}} = \frac{1}{\frac{ε_{0} A}{d} (K_{1} + K_{2})} + \frac{1}{\frac{2 ε_{0} {AK}_{3}}{d}}$
$\frac{1}{C^{''}} = \frac{d}{ε_{0} A (K_{1} + K_{2})} + \frac{d}{2 ε_{0} {AK}_{3}}$
$\frac{1}{C^{''}} = \frac{d}{ε_{0} A} [\frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}]$
Where, $(C^{''} = \frac{ε_{0} AK}{d})$
Then, $\frac{1}{K} = \frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}$

[UPSEE - 2015]

Capacitance

165987 A spherical condenser has inner and outer spheres of radii a and $b$ respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be

1 Zero

2

4 π ε_{0} a

3

4 π ε_{0} b

4

4 π ε_{0} a (\frac{b}{b - a})

Explanation:

: Given, the Radius of inner sphere $= a$ ,
The Radius of outer Sphere $= b$
The expression to calculate the capacitance when outer sphere is earthed.
$C_{1} = 4 π ε_{0} \frac{ab}{b - a}$
The expression to calculate the capacitance when inner sphere is earthed
$C_{2} = 4 π ε_{0} b + \frac{4 π ε_{0} ab}{b - a}$
$C_{2} = 4 π ε_{0} (\frac{b^{2}}{b - a})$
Difference between equation (i) and (ii)
$C_{2} - C_{1} = 4 π ε_{0} (\frac{b^{2}}{b - a}) - 4 π ε_{0} \frac{ab}{b - a}$
$C_{2} - C_{1} = 4 π ε_{0} b [\frac{b - a}{b - a}]$
$C_{2} - C_{1} = 4 π ε_{0} b$

[UPSEE - 2014]

Capacitance

165988 Capacitance of a capacitor made by a thin metal foil is $2 μ F$ . If the foils is folded with paper of thickness $0.15 mm$ , dielectric constant of paper is 2.5 and width of paper is $400 mm$ , the length of the foil will be

1

0.34 m

2

1.33 m

3

13.4 m

4

33.9 m

Explanation:

: Given, $C = 2 \times 10^{- 6} F$
Width of paper $(b) = 400 mm = 0.4 m$
Thickness of paper $(t) = 0.15 mm = 15 \times 10^{- 5} m$
Dielectric constant of paper $(K) = 2.5$
As capacitance for parallel plate capacitor is,
$C = \frac{K ε_{0} A}{d}$
$C = \frac{K ε_{0} (l \times b)}{t}$
$l = \frac{Ct}{K ε_{0} b}$
$l = \frac{2 \times 10^{- 6} \times 15 \times 10^{- 5}}{(2.5 \times 8.85 \times 10^{- 12} \times 0.4)} = 33.89 ⊔ 33.9$
$l = 33.9 m .$

[UPSEE - 2012]

Capacitance

165990 In the condenser shown in the circuit is charged to $5 V$ and left in the circuit, in $12 s$ the charge on the condenser will become :

1

\frac{10}{e} C

2

\frac{e}{10} C

3

\frac{10}{e^{2}} C

4

\frac{e^{2}}{10} C

5

= 2.718

Explanation:

: Given, $C = 2 F, V = 5 V, t = 12 \sec, R = 6 Ω$ Exponential decay of charge take place.
In $C - R$ Circuit, discharging take place.
$q = Q e^{- t / RC}$
Where, $R$ in resistance, $C$ is Capacitance and
$Q = CV$
Hence, $q = CV e^{- t / RC}$
$q = (2 \times 5 \times e^{- 12 / 6 \times 2}) Coulomb$
$q = \frac{10}{e} C$

[UPSEE - 2006

Capacitance

165991 A capacitor having capacitance $1 μ F$ with air is filled with two dielectrics as shown. How many times capacitance will increase?

1 12

2 6

3

8 / 3

4 3

Explanation:

: Given that,
Dielectric constant, $K_{1} = 8, K_{2} = 4$
We know that,
$C = \frac{ε_{0} A}{d}$
$∴ C_{1} = \frac{K_{1} ε_{0} A / 2}{d} = \frac{8}{2} \frac{ε_{0} A}{d} = 4 C$
$C_{2} = \frac{K_{2} ε_{0} A / 2}{d} = \frac{4}{2} \frac{ε_{0} A}{d} = 2 C$
Capacitance will increase due to parallel combination
$C^{'} = C_{1} + C_{2}$
$= 4 C + 2 C$
$= 6 C$

CGPET-2010]

Capacitance

165986 A parallel plate capacitor of area $A$ , plate separation $d$ and capacitance $C$ is filled with three dielectric materials having dielectric constants $k_{1}, k_{2}$ and $k_{3}$ as shown. If a single dielectric material is to be used to have the same capacitance $C$ in the capacitor, then its dielectric constant $k$ is given by

1

K = K_{1} + K_{2} + 2 K_{3}

2

K = \frac{K_{1} K_{2}}{K_{1} + K_{2}} + 2 K_{3}

3

\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{2 K_{3}}

4

\frac{1}{K} = \frac{1}{K_{1} + K_{2}} + \frac{1}{2 K_{3}}

Explanation:

: In the given figure, we can divide it into three different capacitors. Let the capacitance with dielectric constant $K_{1}, K_{2}$ and $K_{3}$ be $C_{1}, C_{2}$ and $C_{3}$ capacity, the width of capacitor $= \frac{d}{2}$ and area is $A / 2$
We have, $C_{1} = \frac{ε_{0} (A / 2) K_{1}}{(d / 2)} = \frac{ε_{0} {AK}_{1}}{d}$
$C_{2} = \frac{ε_{0} (A / 2) K_{2}}{(d / 2)} = \frac{ε_{0} {AK}_{2}}{d}$
$C_{3} = \frac{ε_{0} {AK}_{3}}{(d / 2)} = \frac{2 ε_{0} {AK}_{3}}{d}$
The capacitors $C_{1}$ and $C_{2}$ are in parallel
$∴ C^{'} = C_{1} + C_{2} = \frac{ε_{0} A}{d} (K_{1} + K_{2})$
This combination is in series with $C_{3}$
The net capacitance
$\frac{1}{C^{''}} = \frac{1}{C^{'}} + \frac{1}{C_{3}}$
$\frac{1}{C^{''}} = \frac{1}{\frac{ε_{0} A}{d} (K_{1} + K_{2})} + \frac{1}{\frac{2 ε_{0} {AK}_{3}}{d}}$
$\frac{1}{C^{''}} = \frac{d}{ε_{0} A (K_{1} + K_{2})} + \frac{d}{2 ε_{0} {AK}_{3}}$
$\frac{1}{C^{''}} = \frac{d}{ε_{0} A} [\frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}]$
Where, $(C^{''} = \frac{ε_{0} AK}{d})$
Then, $\frac{1}{K} = \frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}$

[UPSEE - 2015]

Capacitance

165987 A spherical condenser has inner and outer spheres of radii a and $b$ respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be

1 Zero

2

4 π ε_{0} a

3

4 π ε_{0} b

4

4 π ε_{0} a (\frac{b}{b - a})

Explanation:

: Given, the Radius of inner sphere $= a$ ,
The Radius of outer Sphere $= b$
The expression to calculate the capacitance when outer sphere is earthed.
$C_{1} = 4 π ε_{0} \frac{ab}{b - a}$
The expression to calculate the capacitance when inner sphere is earthed
$C_{2} = 4 π ε_{0} b + \frac{4 π ε_{0} ab}{b - a}$
$C_{2} = 4 π ε_{0} (\frac{b^{2}}{b - a})$
Difference between equation (i) and (ii)
$C_{2} - C_{1} = 4 π ε_{0} (\frac{b^{2}}{b - a}) - 4 π ε_{0} \frac{ab}{b - a}$
$C_{2} - C_{1} = 4 π ε_{0} b [\frac{b - a}{b - a}]$
$C_{2} - C_{1} = 4 π ε_{0} b$

[UPSEE - 2014]

Capacitance

165988 Capacitance of a capacitor made by a thin metal foil is $2 μ F$ . If the foils is folded with paper of thickness $0.15 mm$ , dielectric constant of paper is 2.5 and width of paper is $400 mm$ , the length of the foil will be

1

0.34 m

2

1.33 m

3

13.4 m

4

33.9 m

Explanation:

: Given, $C = 2 \times 10^{- 6} F$
Width of paper $(b) = 400 mm = 0.4 m$
Thickness of paper $(t) = 0.15 mm = 15 \times 10^{- 5} m$
Dielectric constant of paper $(K) = 2.5$
As capacitance for parallel plate capacitor is,
$C = \frac{K ε_{0} A}{d}$
$C = \frac{K ε_{0} (l \times b)}{t}$
$l = \frac{Ct}{K ε_{0} b}$
$l = \frac{2 \times 10^{- 6} \times 15 \times 10^{- 5}}{(2.5 \times 8.85 \times 10^{- 12} \times 0.4)} = 33.89 ⊔ 33.9$
$l = 33.9 m .$

[UPSEE - 2012]

Capacitance

165990 In the condenser shown in the circuit is charged to $5 V$ and left in the circuit, in $12 s$ the charge on the condenser will become :

1

\frac{10}{e} C

2

\frac{e}{10} C

3

\frac{10}{e^{2}} C

4

\frac{e^{2}}{10} C

5

= 2.718

Explanation:

: Given, $C = 2 F, V = 5 V, t = 12 \sec, R = 6 Ω$ Exponential decay of charge take place.
In $C - R$ Circuit, discharging take place.
$q = Q e^{- t / RC}$
Where, $R$ in resistance, $C$ is Capacitance and
$Q = CV$
Hence, $q = CV e^{- t / RC}$
$q = (2 \times 5 \times e^{- 12 / 6 \times 2}) Coulomb$
$q = \frac{10}{e} C$

[UPSEE - 2006

Capacitance

165991 A capacitor having capacitance $1 μ F$ with air is filled with two dielectrics as shown. How many times capacitance will increase?

1 12

2 6

3

8 / 3

4 3

Explanation:

: Given that,
Dielectric constant, $K_{1} = 8, K_{2} = 4$
We know that,
$C = \frac{ε_{0} A}{d}$
$∴ C_{1} = \frac{K_{1} ε_{0} A / 2}{d} = \frac{8}{2} \frac{ε_{0} A}{d} = 4 C$
$C_{2} = \frac{K_{2} ε_{0} A / 2}{d} = \frac{4}{2} \frac{ε_{0} A}{d} = 2 C$
Capacitance will increase due to parallel combination
$C^{'} = C_{1} + C_{2}$
$= 4 C + 2 C$
$= 6 C$

CGPET-2010]

Capacitance

165986 A parallel plate capacitor of area $A$ , plate separation $d$ and capacitance $C$ is filled with three dielectric materials having dielectric constants $k_{1}, k_{2}$ and $k_{3}$ as shown. If a single dielectric material is to be used to have the same capacitance $C$ in the capacitor, then its dielectric constant $k$ is given by

1

K = K_{1} + K_{2} + 2 K_{3}

2

K = \frac{K_{1} K_{2}}{K_{1} + K_{2}} + 2 K_{3}

3

\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{2 K_{3}}

4

\frac{1}{K} = \frac{1}{K_{1} + K_{2}} + \frac{1}{2 K_{3}}

Explanation:

: In the given figure, we can divide it into three different capacitors. Let the capacitance with dielectric constant $K_{1}, K_{2}$ and $K_{3}$ be $C_{1}, C_{2}$ and $C_{3}$ capacity, the width of capacitor $= \frac{d}{2}$ and area is $A / 2$
We have, $C_{1} = \frac{ε_{0} (A / 2) K_{1}}{(d / 2)} = \frac{ε_{0} {AK}_{1}}{d}$
$C_{2} = \frac{ε_{0} (A / 2) K_{2}}{(d / 2)} = \frac{ε_{0} {AK}_{2}}{d}$
$C_{3} = \frac{ε_{0} {AK}_{3}}{(d / 2)} = \frac{2 ε_{0} {AK}_{3}}{d}$
The capacitors $C_{1}$ and $C_{2}$ are in parallel
$∴ C^{'} = C_{1} + C_{2} = \frac{ε_{0} A}{d} (K_{1} + K_{2})$
This combination is in series with $C_{3}$
The net capacitance
$\frac{1}{C^{''}} = \frac{1}{C^{'}} + \frac{1}{C_{3}}$
$\frac{1}{C^{''}} = \frac{1}{\frac{ε_{0} A}{d} (K_{1} + K_{2})} + \frac{1}{\frac{2 ε_{0} {AK}_{3}}{d}}$
$\frac{1}{C^{''}} = \frac{d}{ε_{0} A (K_{1} + K_{2})} + \frac{d}{2 ε_{0} {AK}_{3}}$
$\frac{1}{C^{''}} = \frac{d}{ε_{0} A} [\frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}]$
Where, $(C^{''} = \frac{ε_{0} AK}{d})$
Then, $\frac{1}{K} = \frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}$

[UPSEE - 2015]

Capacitance

165987 A spherical condenser has inner and outer spheres of radii a and $b$ respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be

1 Zero

2

4 π ε_{0} a

3

4 π ε_{0} b

4

4 π ε_{0} a (\frac{b}{b - a})

Explanation:

: Given, the Radius of inner sphere $= a$ ,
The Radius of outer Sphere $= b$
The expression to calculate the capacitance when outer sphere is earthed.
$C_{1} = 4 π ε_{0} \frac{ab}{b - a}$
The expression to calculate the capacitance when inner sphere is earthed
$C_{2} = 4 π ε_{0} b + \frac{4 π ε_{0} ab}{b - a}$
$C_{2} = 4 π ε_{0} (\frac{b^{2}}{b - a})$
Difference between equation (i) and (ii)
$C_{2} - C_{1} = 4 π ε_{0} (\frac{b^{2}}{b - a}) - 4 π ε_{0} \frac{ab}{b - a}$
$C_{2} - C_{1} = 4 π ε_{0} b [\frac{b - a}{b - a}]$
$C_{2} - C_{1} = 4 π ε_{0} b$

[UPSEE - 2014]

Capacitance

165988 Capacitance of a capacitor made by a thin metal foil is $2 μ F$ . If the foils is folded with paper of thickness $0.15 mm$ , dielectric constant of paper is 2.5 and width of paper is $400 mm$ , the length of the foil will be

1

0.34 m

2

1.33 m

3

13.4 m

4

33.9 m

Explanation:

: Given, $C = 2 \times 10^{- 6} F$
Width of paper $(b) = 400 mm = 0.4 m$
Thickness of paper $(t) = 0.15 mm = 15 \times 10^{- 5} m$
Dielectric constant of paper $(K) = 2.5$
As capacitance for parallel plate capacitor is,
$C = \frac{K ε_{0} A}{d}$
$C = \frac{K ε_{0} (l \times b)}{t}$
$l = \frac{Ct}{K ε_{0} b}$
$l = \frac{2 \times 10^{- 6} \times 15 \times 10^{- 5}}{(2.5 \times 8.85 \times 10^{- 12} \times 0.4)} = 33.89 ⊔ 33.9$
$l = 33.9 m .$

[UPSEE - 2012]

Capacitance

165990 In the condenser shown in the circuit is charged to $5 V$ and left in the circuit, in $12 s$ the charge on the condenser will become :

1

\frac{10}{e} C

2

\frac{e}{10} C

3

\frac{10}{e^{2}} C

4

\frac{e^{2}}{10} C

5

= 2.718

Explanation:

: Given, $C = 2 F, V = 5 V, t = 12 \sec, R = 6 Ω$ Exponential decay of charge take place.
In $C - R$ Circuit, discharging take place.
$q = Q e^{- t / RC}$
Where, $R$ in resistance, $C$ is Capacitance and
$Q = CV$
Hence, $q = CV e^{- t / RC}$
$q = (2 \times 5 \times e^{- 12 / 6 \times 2}) Coulomb$
$q = \frac{10}{e} C$

[UPSEE - 2006

Capacitance

165991 A capacitor having capacitance $1 μ F$ with air is filled with two dielectrics as shown. How many times capacitance will increase?

1 12

2 6

3

8 / 3

4 3

Explanation:

: Given that,
Dielectric constant, $K_{1} = 8, K_{2} = 4$
We know that,
$C = \frac{ε_{0} A}{d}$
$∴ C_{1} = \frac{K_{1} ε_{0} A / 2}{d} = \frac{8}{2} \frac{ε_{0} A}{d} = 4 C$
$C_{2} = \frac{K_{2} ε_{0} A / 2}{d} = \frac{4}{2} \frac{ε_{0} A}{d} = 2 C$
Capacitance will increase due to parallel combination
$C^{'} = C_{1} + C_{2}$
$= 4 C + 2 C$
$= 6 C$

CGPET-2010]

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: