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Question

(A) Let the two conducting balls be $A$ and $B$ with charges $+q$ and $-q$ respectively. The potential at the surface of ball $A$ due to its own charg

Vedclass · Accepted Answer

$2 \pi \varepsilon_0 r$

Vedclass · Answer

(A) Let the two conducting balls be $A$ and $B$ with charges $+q$ and $-q$ respectively. The potential at the surface of ball $A$ due to its own charge is $V_{A,self} = \frac{kq}{r}$ and due to ball $B$ is $V_{A,ext} = \frac{k(-q)}{d}$.Thus,$V_A = \frac{kq}{r} - \frac{kq}{d}$.Similarly,the potential at the surface of ball $B$ is $V_B = \frac{k(-q)}{r} + \frac{kq}{d} = -\left(\frac{kq}{r} - \frac{kq}{d}\right)$.The potential difference between the two balls is $V = V_A - V_B = 2 \left(\frac{kq}{r} - \frac{kq}{d}\right) = 2kq \left(\frac{1}{r} - \frac{1}{d}\right)$.Since $d \gg r$,we have $\frac{1}{d} \approx 0$,so $V \approx \frac{2kq}{r} = \frac{2q}{4 \pi \varepsilon_0 r} = \frac{q}{2 \pi \varepsilon_0 r}$.The capacitance $C$ is given by $C = \frac{q}{V} = \frac{q}{q / (2 \pi \varepsilon_0 r)} = 2 \pi \varepsilon_0 r$.

Capacitor Type	Capacitance Formula
$A$. Cylindrical capacitor	$i$. $4\pi \epsilon_0 R$
$B$. Spherical capacitor	$ii$. $\frac{K A \epsilon_0}{d}$
$C$. Parallel plate capacitor with dielectric	$iii$. $\frac{2\pi \epsilon_0 \ell}{\ln(r_2/r_1)}$
$D$. Isolated spherical conductor	$iv$. $\frac{4\pi \epsilon_0 r_1 r_2}{r_2 - r_1}$

Two similar conducting balls having charges $+q$ and $-q$ are placed at a separation $d$ from each other in air. The radius of each ball is $r$ and the separation between their centres is $d$ $(d \gg r)$. Calculate the capacitance of the two-ball system.

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