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

Q: 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

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

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$4 \pi \varepsilon_0 r$
B$2 \pi \varepsilon_0$
C$4 \pi \log _e \frac{\varepsilon_0 r}{d}$
D$4 \pi \log _e \frac{r}{d}$

Medium

STD 11 Science
NEET
STD 12 - 2. Electric potential and capacitance
Physics

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