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2. Electric Potential and Capacitance
medium
Two electric charges $12\,\mu C$ and $ - 6\,\mu C$ are placed $20\, cm$ apart in air. There will be a point $P$ on the line joining these charges and outside the region between them, at which the electric potential is zero. The distance of $P$ from $ - 6\,\mu C$ charge is.......$m$
A
$0.10$
B
$0.15 $
C
$0.20 $
D
$0.25$
Solution
(c) Point $P$ will lie near the charge which is smaller in magnitude i.e. $-6\, µ C$. Hence potential at $P$
$V = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{( – 6 \times {{10}^{ – 6}})}}{x} + \frac{1}{{4\pi {\varepsilon _0}}}\frac{{(12 \times {{10}^{ – 6}})}}{{(0.2 + x)}} = 0$
$x = 0.2 \,m$
Standard 12
Physics
