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Capacitance

165605 Spherical capacitors has outer sphere of radius $5 cm$ and inner sphere of radius $2 cm$ . when the inner sphere is earthed, its capacity is $C_{1}$ and when the outer sphere is earthed its capacity is $C_{2}$ . Then $\frac{C_{1}}{C_{2}}$ is

1

\frac{5}{2}

2

\frac{2}{5}

3

\frac{7}{3}

4

\frac{3}{7}

Explanation:

: Given that,
Radius of inner sphere, $R_{1} = 2 cm$
Radius of outer sphere, $R_{2} = 5 cm$
When outer shell is earthed
$C_{2} = 4 π ε_{o} \frac{R_{1} R_{2}}{(R_{2} - R_{1})}$
When inner shell is earthed
Then
$C_{1} = C_{2} + 4 π ε_{o} R_{2}$
$C_{1} = 4 π ε_{o} [\frac{R_{1} R_{2}}{R_{2} - R_{1}} + R_{2}]$
$= 4 π ε_{o} [\frac{2 \times 5}{5 - 2} + 5]$
$C_{1} = 4 π ε_{o} \times \frac{25}{3} Farad$
And, $C_{2} = 4 π ε_{o} \frac{R_{1} R_{2}}{(R_{2} - R_{1})}$
$= 4 π ε_{o} \frac{2 \times 5}{(5 - 2)} = 4 π ε_{o} \frac{10}{3} Farad$
Hence, $\frac{C_{1}}{C_{2}} = \frac{25 / 3}{10 / 3} = 5 / 2$

[AP EAMCET (23.04.2019) Shift-I]

Capacitance

165606 27 small drops each having charge $q$ and radius $r$ coalesce to form big drop. How many times charge and capacitance will become?

1 3,27

2 27,3

3 27,27

4 3,3

Explanation:

: Given,
Total number of drop $= 27$
Radius of big drop $= R$
Radius of small drop $= r$
In coalescing into a single drop, charge remains conserved charge on big drop $= 27 \times$ Charge of small drop
$q^{'} = 27 q$
Since, Volume is also remain same.
$\frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3}$
$R^{3} = (3 r)^{3}$
$R = 3 r$
So, capacitance of small drop $= 4 π ε_{0} r$
And, capacitance of bigger drop
$C^{'} = 4 π ε_{0} R$
$= 4 π ε_{0} (3 r)$
$= 12 π ε_{0} r = 3 \times (4 π ε_{0} r)$
Hence, $C^{'} = 3 C$

[JCECE-2007

Capacitance

165607 Figure below shows four plates each of area $A$ and separated from one another by a distance $d$ . What is the capacitance between $P$ and $Q$ ?

1

\frac{ε_{0} A}{d}

2

\frac{2 ε_{0} A}{d}

3

\frac{3 ε_{0} A}{d}

4

\frac{4 ε_{0} A}{d}

5 Zero

Explanation:

: From question,
The given circuit is equivalent to a parallel combination two identical capacitors.
Hence, equivalent capacitance between $P$ and $Q$ is.

Two capacitors $C$ in parallel order are formed between the plates.
$C_{PQ} = C_{1} + C_{2}$
$C_{PQ} = \frac{ε_{0} A}{d} + \frac{ε_{0} A}{d} = \frac{2 ε_{0} A}{d}$

[WBJEE-2013

Capacitance

165608 The capacitance of a spherical conductor with radius $1 m$ is

1

9 \times 10^{9} F

2

1 μ F

3

1.1 \times 10^{- 10} F

4

1 \times 10^{- 6} F

Explanation:

: Given,
Radius of spherical conductor $(r) = 1 m$
We know the capacitance of spherical conductor -
$C = 4 π ε_{0} r$
$C = \frac{r}{\frac{1}{4 π ε_{0}}}$
Since, $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} C^{2}$
So,
$C = \frac{1}{9 \times 10^{9}} {r = 1 m}$
$C = 0.11 \times 10^{- 9} F$
$C = 1.1 \times 10^{- 10} F$

[JCECE-2014]

Capacitance

165609 In the figure, charge and the potential difference across the $4 μ F$ capacitor will be nearly

1

600 μ C, 150 V

2

300 μ C, 75 V

3

800 μ C, 200 V

4

580 μ C, 145 V

Explanation:

: Given diagram,

From figure,
All capacitor in series,
$\frac{1}{C_{eq}} = \frac{1}{8} + \frac{1}{20} + \frac{1}{12} = \frac{31}{120}$
$C_{eq} = \frac{120}{31} μ F$
Total charge of $(8 μ F) = Q = CV$
$Q = \frac{120}{31} \times 300 = 1161 μ C$
Then, charge of $(4 μ F) = \frac{Q}{2} = \frac{1161}{2} μ C$
$= 580 μ C$
Hence, Potential difference, $(V) = \frac{Q}{C}$
$= \frac{580}{4} = 145 V$

[JCECE-2011]

Capacitance

165605 Spherical capacitors has outer sphere of radius $5 cm$ and inner sphere of radius $2 cm$ . when the inner sphere is earthed, its capacity is $C_{1}$ and when the outer sphere is earthed its capacity is $C_{2}$ . Then $\frac{C_{1}}{C_{2}}$ is

1

\frac{5}{2}

2

\frac{2}{5}

3

\frac{7}{3}

4

\frac{3}{7}

Explanation:

: Given that,
Radius of inner sphere, $R_{1} = 2 cm$
Radius of outer sphere, $R_{2} = 5 cm$
When outer shell is earthed
$C_{2} = 4 π ε_{o} \frac{R_{1} R_{2}}{(R_{2} - R_{1})}$
When inner shell is earthed
Then
$C_{1} = C_{2} + 4 π ε_{o} R_{2}$
$C_{1} = 4 π ε_{o} [\frac{R_{1} R_{2}}{R_{2} - R_{1}} + R_{2}]$
$= 4 π ε_{o} [\frac{2 \times 5}{5 - 2} + 5]$
$C_{1} = 4 π ε_{o} \times \frac{25}{3} Farad$
And, $C_{2} = 4 π ε_{o} \frac{R_{1} R_{2}}{(R_{2} - R_{1})}$
$= 4 π ε_{o} \frac{2 \times 5}{(5 - 2)} = 4 π ε_{o} \frac{10}{3} Farad$
Hence, $\frac{C_{1}}{C_{2}} = \frac{25 / 3}{10 / 3} = 5 / 2$

[AP EAMCET (23.04.2019) Shift-I]

Capacitance

165606 27 small drops each having charge $q$ and radius $r$ coalesce to form big drop. How many times charge and capacitance will become?

1 3,27

2 27,3

3 27,27

4 3,3

Explanation:

: Given,
Total number of drop $= 27$
Radius of big drop $= R$
Radius of small drop $= r$
In coalescing into a single drop, charge remains conserved charge on big drop $= 27 \times$ Charge of small drop
$q^{'} = 27 q$
Since, Volume is also remain same.
$\frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3}$
$R^{3} = (3 r)^{3}$
$R = 3 r$
So, capacitance of small drop $= 4 π ε_{0} r$
And, capacitance of bigger drop
$C^{'} = 4 π ε_{0} R$
$= 4 π ε_{0} (3 r)$
$= 12 π ε_{0} r = 3 \times (4 π ε_{0} r)$
Hence, $C^{'} = 3 C$

[JCECE-2007

Capacitance

165607 Figure below shows four plates each of area $A$ and separated from one another by a distance $d$ . What is the capacitance between $P$ and $Q$ ?

1

\frac{ε_{0} A}{d}

2

\frac{2 ε_{0} A}{d}

3

\frac{3 ε_{0} A}{d}

4

\frac{4 ε_{0} A}{d}

5 Zero

Explanation:

: From question,
The given circuit is equivalent to a parallel combination two identical capacitors.
Hence, equivalent capacitance between $P$ and $Q$ is.

Two capacitors $C$ in parallel order are formed between the plates.
$C_{PQ} = C_{1} + C_{2}$
$C_{PQ} = \frac{ε_{0} A}{d} + \frac{ε_{0} A}{d} = \frac{2 ε_{0} A}{d}$

[WBJEE-2013

Capacitance

165608 The capacitance of a spherical conductor with radius $1 m$ is

1

9 \times 10^{9} F

2

1 μ F

3

1.1 \times 10^{- 10} F

4

1 \times 10^{- 6} F

Explanation:

: Given,
Radius of spherical conductor $(r) = 1 m$
We know the capacitance of spherical conductor -
$C = 4 π ε_{0} r$
$C = \frac{r}{\frac{1}{4 π ε_{0}}}$
Since, $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} C^{2}$
So,
$C = \frac{1}{9 \times 10^{9}} {r = 1 m}$
$C = 0.11 \times 10^{- 9} F$
$C = 1.1 \times 10^{- 10} F$

[JCECE-2014]

Capacitance

165609 In the figure, charge and the potential difference across the $4 μ F$ capacitor will be nearly

1

600 μ C, 150 V

2

300 μ C, 75 V

3

800 μ C, 200 V

4

580 μ C, 145 V

Explanation:

: Given diagram,

From figure,
All capacitor in series,
$\frac{1}{C_{eq}} = \frac{1}{8} + \frac{1}{20} + \frac{1}{12} = \frac{31}{120}$
$C_{eq} = \frac{120}{31} μ F$
Total charge of $(8 μ F) = Q = CV$
$Q = \frac{120}{31} \times 300 = 1161 μ C$
Then, charge of $(4 μ F) = \frac{Q}{2} = \frac{1161}{2} μ C$
$= 580 μ C$
Hence, Potential difference, $(V) = \frac{Q}{C}$
$= \frac{580}{4} = 145 V$

[JCECE-2011]

Capacitance

165605 Spherical capacitors has outer sphere of radius $5 cm$ and inner sphere of radius $2 cm$ . when the inner sphere is earthed, its capacity is $C_{1}$ and when the outer sphere is earthed its capacity is $C_{2}$ . Then $\frac{C_{1}}{C_{2}}$ is

1

\frac{5}{2}

2

\frac{2}{5}

3

\frac{7}{3}

4

\frac{3}{7}

Explanation:

: Given that,
Radius of inner sphere, $R_{1} = 2 cm$
Radius of outer sphere, $R_{2} = 5 cm$
When outer shell is earthed
$C_{2} = 4 π ε_{o} \frac{R_{1} R_{2}}{(R_{2} - R_{1})}$
When inner shell is earthed
Then
$C_{1} = C_{2} + 4 π ε_{o} R_{2}$
$C_{1} = 4 π ε_{o} [\frac{R_{1} R_{2}}{R_{2} - R_{1}} + R_{2}]$
$= 4 π ε_{o} [\frac{2 \times 5}{5 - 2} + 5]$
$C_{1} = 4 π ε_{o} \times \frac{25}{3} Farad$
And, $C_{2} = 4 π ε_{o} \frac{R_{1} R_{2}}{(R_{2} - R_{1})}$
$= 4 π ε_{o} \frac{2 \times 5}{(5 - 2)} = 4 π ε_{o} \frac{10}{3} Farad$
Hence, $\frac{C_{1}}{C_{2}} = \frac{25 / 3}{10 / 3} = 5 / 2$

[AP EAMCET (23.04.2019) Shift-I]

Capacitance

165606 27 small drops each having charge $q$ and radius $r$ coalesce to form big drop. How many times charge and capacitance will become?

1 3,27

2 27,3

3 27,27

4 3,3

Explanation:

: Given,
Total number of drop $= 27$
Radius of big drop $= R$
Radius of small drop $= r$
In coalescing into a single drop, charge remains conserved charge on big drop $= 27 \times$ Charge of small drop
$q^{'} = 27 q$
Since, Volume is also remain same.
$\frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3}$
$R^{3} = (3 r)^{3}$
$R = 3 r$
So, capacitance of small drop $= 4 π ε_{0} r$
And, capacitance of bigger drop
$C^{'} = 4 π ε_{0} R$
$= 4 π ε_{0} (3 r)$
$= 12 π ε_{0} r = 3 \times (4 π ε_{0} r)$
Hence, $C^{'} = 3 C$

[JCECE-2007

Capacitance

165607 Figure below shows four plates each of area $A$ and separated from one another by a distance $d$ . What is the capacitance between $P$ and $Q$ ?

1

\frac{ε_{0} A}{d}

2

\frac{2 ε_{0} A}{d}

3

\frac{3 ε_{0} A}{d}

4

\frac{4 ε_{0} A}{d}

5 Zero

Explanation:

: From question,
The given circuit is equivalent to a parallel combination two identical capacitors.
Hence, equivalent capacitance between $P$ and $Q$ is.

Two capacitors $C$ in parallel order are formed between the plates.
$C_{PQ} = C_{1} + C_{2}$
$C_{PQ} = \frac{ε_{0} A}{d} + \frac{ε_{0} A}{d} = \frac{2 ε_{0} A}{d}$

[WBJEE-2013

Capacitance

165608 The capacitance of a spherical conductor with radius $1 m$ is

1

9 \times 10^{9} F

2

1 μ F

3

1.1 \times 10^{- 10} F

4

1 \times 10^{- 6} F

Explanation:

: Given,
Radius of spherical conductor $(r) = 1 m$
We know the capacitance of spherical conductor -
$C = 4 π ε_{0} r$
$C = \frac{r}{\frac{1}{4 π ε_{0}}}$
Since, $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} C^{2}$
So,
$C = \frac{1}{9 \times 10^{9}} {r = 1 m}$
$C = 0.11 \times 10^{- 9} F$
$C = 1.1 \times 10^{- 10} F$

[JCECE-2014]

Capacitance

165609 In the figure, charge and the potential difference across the $4 μ F$ capacitor will be nearly

1

600 μ C, 150 V

2

300 μ C, 75 V

3

800 μ C, 200 V

4

580 μ C, 145 V

Explanation:

: Given diagram,

From figure,
All capacitor in series,
$\frac{1}{C_{eq}} = \frac{1}{8} + \frac{1}{20} + \frac{1}{12} = \frac{31}{120}$
$C_{eq} = \frac{120}{31} μ F$
Total charge of $(8 μ F) = Q = CV$
$Q = \frac{120}{31} \times 300 = 1161 μ C$
Then, charge of $(4 μ F) = \frac{Q}{2} = \frac{1161}{2} μ C$
$= 580 μ C$
Hence, Potential difference, $(V) = \frac{Q}{C}$
$= \frac{580}{4} = 145 V$

[JCECE-2011]

Capacitance

165605 Spherical capacitors has outer sphere of radius $5 cm$ and inner sphere of radius $2 cm$ . when the inner sphere is earthed, its capacity is $C_{1}$ and when the outer sphere is earthed its capacity is $C_{2}$ . Then $\frac{C_{1}}{C_{2}}$ is

1

\frac{5}{2}

2

\frac{2}{5}

3

\frac{7}{3}

4

\frac{3}{7}

Explanation:

: Given that,
Radius of inner sphere, $R_{1} = 2 cm$
Radius of outer sphere, $R_{2} = 5 cm$
When outer shell is earthed
$C_{2} = 4 π ε_{o} \frac{R_{1} R_{2}}{(R_{2} - R_{1})}$
When inner shell is earthed
Then
$C_{1} = C_{2} + 4 π ε_{o} R_{2}$
$C_{1} = 4 π ε_{o} [\frac{R_{1} R_{2}}{R_{2} - R_{1}} + R_{2}]$
$= 4 π ε_{o} [\frac{2 \times 5}{5 - 2} + 5]$
$C_{1} = 4 π ε_{o} \times \frac{25}{3} Farad$
And, $C_{2} = 4 π ε_{o} \frac{R_{1} R_{2}}{(R_{2} - R_{1})}$
$= 4 π ε_{o} \frac{2 \times 5}{(5 - 2)} = 4 π ε_{o} \frac{10}{3} Farad$
Hence, $\frac{C_{1}}{C_{2}} = \frac{25 / 3}{10 / 3} = 5 / 2$

[AP EAMCET (23.04.2019) Shift-I]

Capacitance

165606 27 small drops each having charge $q$ and radius $r$ coalesce to form big drop. How many times charge and capacitance will become?

1 3,27

2 27,3

3 27,27

4 3,3

Explanation:

: Given,
Total number of drop $= 27$
Radius of big drop $= R$
Radius of small drop $= r$
In coalescing into a single drop, charge remains conserved charge on big drop $= 27 \times$ Charge of small drop
$q^{'} = 27 q$
Since, Volume is also remain same.
$\frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3}$
$R^{3} = (3 r)^{3}$
$R = 3 r$
So, capacitance of small drop $= 4 π ε_{0} r$
And, capacitance of bigger drop
$C^{'} = 4 π ε_{0} R$
$= 4 π ε_{0} (3 r)$
$= 12 π ε_{0} r = 3 \times (4 π ε_{0} r)$
Hence, $C^{'} = 3 C$

[JCECE-2007

Capacitance

165607 Figure below shows four plates each of area $A$ and separated from one another by a distance $d$ . What is the capacitance between $P$ and $Q$ ?

1

\frac{ε_{0} A}{d}

2

\frac{2 ε_{0} A}{d}

3

\frac{3 ε_{0} A}{d}

4

\frac{4 ε_{0} A}{d}

5 Zero

Explanation:

: From question,
The given circuit is equivalent to a parallel combination two identical capacitors.
Hence, equivalent capacitance between $P$ and $Q$ is.

Two capacitors $C$ in parallel order are formed between the plates.
$C_{PQ} = C_{1} + C_{2}$
$C_{PQ} = \frac{ε_{0} A}{d} + \frac{ε_{0} A}{d} = \frac{2 ε_{0} A}{d}$

[WBJEE-2013

Capacitance

165608 The capacitance of a spherical conductor with radius $1 m$ is

1

9 \times 10^{9} F

2

1 μ F

3

1.1 \times 10^{- 10} F

4

1 \times 10^{- 6} F

Explanation:

: Given,
Radius of spherical conductor $(r) = 1 m$
We know the capacitance of spherical conductor -
$C = 4 π ε_{0} r$
$C = \frac{r}{\frac{1}{4 π ε_{0}}}$
Since, $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} C^{2}$
So,
$C = \frac{1}{9 \times 10^{9}} {r = 1 m}$
$C = 0.11 \times 10^{- 9} F$
$C = 1.1 \times 10^{- 10} F$

[JCECE-2014]

Capacitance

165609 In the figure, charge and the potential difference across the $4 μ F$ capacitor will be nearly

1

600 μ C, 150 V

2

300 μ C, 75 V

3

800 μ C, 200 V

4

580 μ C, 145 V

Explanation:

: Given diagram,

From figure,
All capacitor in series,
$\frac{1}{C_{eq}} = \frac{1}{8} + \frac{1}{20} + \frac{1}{12} = \frac{31}{120}$
$C_{eq} = \frac{120}{31} μ F$
Total charge of $(8 μ F) = Q = CV$
$Q = \frac{120}{31} \times 300 = 1161 μ C$
Then, charge of $(4 μ F) = \frac{Q}{2} = \frac{1161}{2} μ C$
$= 580 μ C$
Hence, Potential difference, $(V) = \frac{Q}{C}$
$= \frac{580}{4} = 145 V$

[JCECE-2011]

Capacitance

165605 Spherical capacitors has outer sphere of radius $5 cm$ and inner sphere of radius $2 cm$ . when the inner sphere is earthed, its capacity is $C_{1}$ and when the outer sphere is earthed its capacity is $C_{2}$ . Then $\frac{C_{1}}{C_{2}}$ is

1

\frac{5}{2}

2

\frac{2}{5}

3

\frac{7}{3}

4

\frac{3}{7}

Explanation:

: Given that,
Radius of inner sphere, $R_{1} = 2 cm$
Radius of outer sphere, $R_{2} = 5 cm$
When outer shell is earthed
$C_{2} = 4 π ε_{o} \frac{R_{1} R_{2}}{(R_{2} - R_{1})}$
When inner shell is earthed
Then
$C_{1} = C_{2} + 4 π ε_{o} R_{2}$
$C_{1} = 4 π ε_{o} [\frac{R_{1} R_{2}}{R_{2} - R_{1}} + R_{2}]$
$= 4 π ε_{o} [\frac{2 \times 5}{5 - 2} + 5]$
$C_{1} = 4 π ε_{o} \times \frac{25}{3} Farad$
And, $C_{2} = 4 π ε_{o} \frac{R_{1} R_{2}}{(R_{2} - R_{1})}$
$= 4 π ε_{o} \frac{2 \times 5}{(5 - 2)} = 4 π ε_{o} \frac{10}{3} Farad$
Hence, $\frac{C_{1}}{C_{2}} = \frac{25 / 3}{10 / 3} = 5 / 2$

[AP EAMCET (23.04.2019) Shift-I]

Capacitance

165606 27 small drops each having charge $q$ and radius $r$ coalesce to form big drop. How many times charge and capacitance will become?

1 3,27

2 27,3

3 27,27

4 3,3

Explanation:

: Given,
Total number of drop $= 27$
Radius of big drop $= R$
Radius of small drop $= r$
In coalescing into a single drop, charge remains conserved charge on big drop $= 27 \times$ Charge of small drop
$q^{'} = 27 q$
Since, Volume is also remain same.
$\frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3}$
$R^{3} = (3 r)^{3}$
$R = 3 r$
So, capacitance of small drop $= 4 π ε_{0} r$
And, capacitance of bigger drop
$C^{'} = 4 π ε_{0} R$
$= 4 π ε_{0} (3 r)$
$= 12 π ε_{0} r = 3 \times (4 π ε_{0} r)$
Hence, $C^{'} = 3 C$

[JCECE-2007

Capacitance

165607 Figure below shows four plates each of area $A$ and separated from one another by a distance $d$ . What is the capacitance between $P$ and $Q$ ?

1

\frac{ε_{0} A}{d}

2

\frac{2 ε_{0} A}{d}

3

\frac{3 ε_{0} A}{d}

4

\frac{4 ε_{0} A}{d}

5 Zero

Explanation:

: From question,
The given circuit is equivalent to a parallel combination two identical capacitors.
Hence, equivalent capacitance between $P$ and $Q$ is.

Two capacitors $C$ in parallel order are formed between the plates.
$C_{PQ} = C_{1} + C_{2}$
$C_{PQ} = \frac{ε_{0} A}{d} + \frac{ε_{0} A}{d} = \frac{2 ε_{0} A}{d}$

[WBJEE-2013

Capacitance

165608 The capacitance of a spherical conductor with radius $1 m$ is

1

9 \times 10^{9} F

2

1 μ F

3

1.1 \times 10^{- 10} F

4

1 \times 10^{- 6} F

Explanation:

: Given,
Radius of spherical conductor $(r) = 1 m$
We know the capacitance of spherical conductor -
$C = 4 π ε_{0} r$
$C = \frac{r}{\frac{1}{4 π ε_{0}}}$
Since, $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} C^{2}$
So,
$C = \frac{1}{9 \times 10^{9}} {r = 1 m}$
$C = 0.11 \times 10^{- 9} F$
$C = 1.1 \times 10^{- 10} F$

[JCECE-2014]

Capacitance

165609 In the figure, charge and the potential difference across the $4 μ F$ capacitor will be nearly

1

600 μ C, 150 V

2

300 μ C, 75 V

3

800 μ C, 200 V

4

580 μ C, 145 V

Explanation:

: Given diagram,

From figure,
All capacitor in series,
$\frac{1}{C_{eq}} = \frac{1}{8} + \frac{1}{20} + \frac{1}{12} = \frac{31}{120}$
$C_{eq} = \frac{120}{31} μ F$
Total charge of $(8 μ F) = Q = CV$
$Q = \frac{120}{31} \times 300 = 1161 μ C$
Then, charge of $(4 μ F) = \frac{Q}{2} = \frac{1161}{2} μ C$
$= 580 μ C$
Hence, Potential difference, $(V) = \frac{Q}{C}$
$= \frac{580}{4} = 145 V$

[JCECE-2011]