Question

Two charges, each equal to $1\mu C$, are placed at the vertices $A$ and $B$ of a triangle $ABC$. The product of $AC$ and $BC$ is $30c{m}^2.$ The sum of the sides $AC$ and $BC$ is $10cm$. The potential at $C$ is

• A. $3\times 10^{5}V$
• B. $6\times 10^{5}V$
• C. $9\times 10^{5}V$
• D. 18 m V

Answer 1

Answer: (A)

$V=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q}{AC}+\frac{q}{BC}\right]$
$=\frac{q}{4\pi {\epsilon }_0}\left[\frac{AC+BC}{AC\times BC}\right]$
or $V=\frac{9\times 10^9\times 1\times 10^{-6}}{1}\times \left[\frac{10}{30}\times 100\right]$ volt
$=3\times 10^5$ volt