Two similar conducting balls having charges + | vedclass

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Question

(b)

$V_A=\frac{k q}{r}+\frac{k(-q)}{d-r}$

$V_B=\frac{-k q}{r}+\frac{k q}{d-r}$

$V=V_A-V_B=\frac{2 k q}{r}-\frac{2 k q}{d-r}=\frac{2 q}{4 \pi

Vedclass · Accepted Answer

$2 \pi \varepsilon_0$

Vedclass · Answer

(b)

$V_A=\frac{k q}{r}+\frac{k(-q)}{d-r}$

$V_B=\frac{-k q}{r}+\frac{k q}{d-r}$

$V=V_A-V_B=\frac{2 k q}{r}-\frac{2 k q}{d-r}=\frac{2 q}{4 \pi \varepsilon_0}\left[\frac{1}{r}-\frac{1}{d-r}\right]$

$V=\frac{q}{2 \pi \varepsilon_0}\left[\frac{d-2 r}{(r)(d-r)}\right]$

$d \gg r$

$V=\frac{q}{2 \pi \varepsilon_0}$

$\frac{q}{V}=2 \pi \varepsilon_0 r=C$

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 $d(d > > r)$. Calculate the capacitance of the two ball system

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