Capacitance of isolated bodies - class-XII

${C} _{1}=\cfrac{{ \varepsilon  } _{ 0 }A}{d}$ and ${C} _{2}=\cfrac{K{ \varepsilon  } _{ 0 }A}{2d}$
$\Rightarrow$ $\cfrac{{C} _{2}}{{C} _{1}}=\cfrac{K}{2}=\cfrac{40\times {10}^{-12}}{10\times {10}^{-12}}$
$\Rightarrow$ $K=8$

Because earth can store infinite charge so its capacitance is infinite, therefore it is at a zero potential. 

Vol. of the big drop =1000 *vol. of each small drop
$\displaystyle \frac{4}{3} \pi R^3 = 1000\times \frac{4}{3} \pi r^3$
$\Rightarrow R^3 = (10r)^3$
$\Rightarrow $ R =10r
Potential of each charged sphere for small drop, V = $\displaystyle \frac{q}{c}$
$\therefore$ Potential for the big drop = $1000 \displaystyle \frac{q}{C}$
We have, $C= 4 \pi \epsilon _0 R = 4 \pi \epsilon _0 \times 10 r$
$\therefore$ Potential of the big drop 
=$\displaystyle \frac{1000}{10} . \frac{q}{4 \pi \epsilon _0 r} =100V$

Given,

Electric field E = -dV/dy = -(54.8 - 56) / (12.5 - 12.3) = -(-1.2) / 0.2 = 6 V/cm = 600 V/m. Force F = qE = 4 * 10^-6 * 600 = 24 * 10^-4 N.

The capacity of an isolated spherical conductor of radius $R$ is $4\pi \epsilon _{0}R$
$\therefore C\propto R$.

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

For three 6 uF capacitors: Series combination C_min = 6/3 = 2 uF. Parallel combination C_max = 6 * 3 = 18 uF.

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

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

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

C = epsilon_0 * A / d. A = C * d / epsilon_0 = 2 * 0.005 / (8.85 * 10^-12) = 1.13 * 10^9 m^2 = 1130 km^2.

Capacitance of isolated sphere C1 = 4 * pi * epsilon_0 * a. Capacitance of spherical capacitor C2 = 4 * pi * epsilon_0 * a * b / (b - a). Given b/a = n/(n-1), then b = a * n / (n-1). Substituting gives C2 = C1 * n.

Capacitance is given as

As per definition of capacitance 

According to Gauss's Law, the electric field inside a hollow charged conductor is zero because there is no enclosed charge.