Tag: capacitance of isolated bodies

Questions Related to capacitance of isolated bodies

Two metal spheres of capacitance, ${C} _{1}$ and ${C} _{2}$ carry some charges. They are put in contact and then separated. The final charges ${Q} _{1}$ and ${Q} _{2}$ on them will satisfy:

capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

  1. $\dfrac { { Q } _{ 1 } }{ { Q } _{ 2 } } <\dfrac { { C } _{ 1 } }{ { C } _{ 2 } }$

  2. $\dfrac { { Q } _{ 1 } }{ { Q } _{ 2 } } =\dfrac { { C } _{ 1 } }{ { C } _{ 2 } }$

  3. $\dfrac { { Q } _{ 1 } }{ { Q } _{ 2 } } >\dfrac { { C } _{ 1 } }{ { C } _{ 2 } }$

  4. $\dfrac { { Q } _{ 1 } }{ { Q } _{ 2 } } =\dfrac { { C } _{ 2 } }{ { C } _{ 1 } }$

Show answer
Correct Option: B
Explanation:

Let the charge on two sphere initially are $q _1\ &amp;\ q _2$. Now when these two capacitors (spheres) are kept in contact with each other and separated. Then charges on the two spheres are,


Let $Q _1\ &amp;\ Q _2$ are the final charges on spheres. So, final charge will be conserved.
$Q _1+Q _2=q _1+q _2$

$\dfrac{Q _1}{Q _2}=\dfrac{C _1V _1}{C _2V _2}$

The charge will flow until the potential of both the spheres becomes the same.
$\dfrac{Q _1}{Q _2}=\dfrac{C _1V}{C _2V}$

$\dfrac{Q _1}{Q _2}=\dfrac{C _1}{C _2}$

A $110V. 60W$ lamp is run from a $220V$ AC mains using a capacitor in series with the lamp, instead of a resistor then the voltage across the capacitor is about:

capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

  1. $110V$

  2. $190V$

  3. $220V$

  4. $311V$

Show answer
Correct Option: B
Explanation:
$V _c=\sqrt{V^2-V _R^2}$

$=\sqrt{220^2-110^2}$

$=190V$

Three capacitors of capacitances 6 µF each are available. The minimum and maximum capacitances, which may be obtained are

capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

  1. $ 2 \mu F $ and $ 18 \mu F $

  2. $ 5 \mu F $ and $ 5 \mu F $

  3. $ 7 \mu F $ and $ 3 \mu F $

  4. $ 8 \mu F $ and $ 2 \mu F $

Show answer
Correct Option: A

The capacitance of a spherical condenser is $1mF$. If the spacing between the two spheres is $1mm$, then the radius of the outer sphere is

capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

  1. $30cm$

  2. $6m$

  3. $5cm$

  4. $3m$

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} From\, \, the\, question \\ C=\dfrac { { 4\pi { \varepsilon _{ 0 } } } }{ { \left[ { \dfrac { 1 }{ { { r _{ in } } } } -\dfrac { 1 }{ { { r _{ out } } } }  } \right]  } } =\dfrac { { 4\pi \varepsilon  } }{ { \left[ { \dfrac { 1 }{ { { r _{ 1 } } } } -\dfrac { 1 }{ { { r _{ 1 } }+0.001 } }  } \right]  } }  \\ 1\times { 10^{ -6 } }=\dfrac { { 4\times 3.14\times 8.854\times { { 10 }^{ -12 } } } }{ { \left[ { \dfrac { 1 }{ { { r _{ 1 } } } } -\dfrac { 1 }{ { { r _{ 1 } }+0.001 } }  } \right]  } }  \\ \left[ { \dfrac { 1 }{ { { r _{ 1 } } } } -\dfrac { 1 }{ { { r _{ 1 } }+0.001 } }  } \right] =\dfrac { { 4\times 3.14\times 8.854\times { { 10 }^{ -12 } } } }{ { 1\times { { 10 }^{ -6 } } } }  \\ \dfrac { { \left[ { \left( { { r _{ 1 } }+0.001 } \right) -{ r _{ 1 } } } \right]  } }{ { { r _{ 1 } }\times \left( { { r _{ 1 } }+0.001 } \right)  } } =4\times 3.14\times 8.854\times { 10^{ -6 } } \\ { r _{ 1 } }\times \left( { { r _{ 1 } }+0.001 } \right) =\dfrac { { 4\times 3.14\times 8.854\times { { 10 }^{ -6 } } } }{ { 0.001 } }  \\ r _{ 1 }^{ 2 }+0.001{ r _{ 1 } }-4\times 3.14\times 8.854\times { 10^{ -3 } }=0 \\ r _{ 1 }^{ 2 }+0.001{ r _{ 1 } }-0.1112=0 \\ { r _{ 1 } }=0.333m\, \, \, or\, \, { r _{ 1 } }=-334m \\ Since,\, it\, cannot\, be\, negative \\ Thereforem\, radius\, \, of\, outer\, \, sphere\, ={ r _{ 1 } }+0.001 \\ { r _{ outer } }=0.334m \\ or,\, { r _{ 1 } }=33.4cm \\  \end{array}$

Hence, the option $A$ is the correct answer.

The capacitance of a spherical condenser is $1mF$. If the spacing between the two spheres is $1mm$, then the radius of the outer space is

capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

  1. $30cm$

  2. $6m$

  3. $5cm$

  4. $3m$

Show answer
Correct Option: D
Explanation:

$\begin{array}{l} C=\frac { { 4\pi { E _{ 0 } } } }{ { \left( { \frac { 1 }{ { { r _{ i } } } }  } \right) -\left( { \frac { 1 }{ { { r _{ 0 } } } }  } \right)  } } .....................\left( 1 \right)  \ According\, \, to\, \, the\, \, question:- \ { r _{ 0 } }-{ r _{ 1 } }=0.001\, m \ C=0.00000\, 1F..............\left( 2 \right)  \ Putting\, \, \left( 2 \right) \, \, in\, \, \, \left( 1 \right)  \ \therefore r _{ 0 }^{ 2 }-{ r _{ 0 } }\left( { 0.001 } \right) -\left( { 9\times 000000000\times 0.000000001 } \right) =0 \ therefore\, \, { r _{ 0 } }=3\, \, m \end{array}$

A capacitor has capacitance $2F$. plate separation $0.5 cm $ then area of plate  [You will realize from your answer why ordinary capacitors are in the range of μF or less. However, electrolytic capacitors do have a much larger capacitance $(0.1 F)$ because of very minute separation between the conductors.]:

capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

  1. $1130cm^2$

  2. $1130m^2$

  3. $1130km^2$

  4. None of these

Show answer
Correct Option: C

A coil, a capacitor and an A. C. source of rms voltage 24 V are connected in series. By varying the frequency of the source, a maximum rms current of 6 A is observed. If the coil is connected to a battery of emf 12 V and internal resistance $4\Omega$, the current through it will be    

capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

  1. 2.4 A

  2. 1.8 A

  3. 1.5 A

  4. 1.2 A

Show answer
Correct Option: C
Explanation:

$\begin{array}{l} { E _{ rms } }=24V \ r=4\Omega ,\, \, \, { I _{ rms } }=6A \ R=\frac { E }{ I } =\frac { { 24 } }{ 6 } =4\Omega  \ Internal\, { { Re } }sis\tan  ce=4\Omega  \ Hence,\, net\, resis\tan  ce=4+4=8\Omega  \ \therefore Current=\frac { { 12 } }{ 8 } =1.5A \  \end{array}$

Hence, the option $C$ is the correct answer.

The capacitance (C) for an isolated conducting sphere of radius(a) is given by $4\pi \varepsilon _0a$. If the sphere is enclosed with an earthed concentric sphere, the ratio of the radii of the spheres being $\dfrac{n}{(n-1)}$ then the capacitance of such a sphere will be increased by a factor?

capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

  1. $n$

  2. $\dfrac{n}{(n-1)}$

  3. $\dfrac{(n-1)}{n}$

  4. $an$

Show answer
Correct Option: A

If the circumferences of a sphere is $2\ m$, then capacitance of sphere in water would be:

capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

  1. $2700\ pF$

  2. $2760\ pF$

  3. $2780\ pF$

  4. $2846\ pF$

Show answer
Correct Option: D
Explanation:

Capacitance is given as

$C=\varepsilon _0\frac{A}{d}$
For a sphere placed in water, the capacitance will be,
$C=4\pi \varepsilon R$
Here, $\varepsilon$ os the permittivity of water 
In terms of permittivity of free space and dielectric constant of water, we get 
$C=4\pi \varepsilon _0kR$
It is given that circumference is 2m
Hence, $c=2\pi R$  
$\therefore R=\frac{1}{\pi}$
$C=4\pi \varepsilon _0k\frac{1}{\pi}=4\varepsilon _0k$
$C=4\times 8.85\times10^{-12}\times80.4$
$C=2846\times 10^{-12}F$
$C=2846 pF$

If 'Q' is the quantity of charge, 'V' the potential and 'C' the capacity of a conductor, they are related as:

capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

  1. $C = QV$

  2. $Q = VC$

  3. $V = CQ$

  4. $CVQ = constant$

Show answer
Correct Option: B
Explanation:

As per definition of capacitance 

$C=\dfrac{Q}{V}$
SO $Q=VC$