Question

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

$V=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{A C}+\frac{q}{B C}\right]$
$=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{A C+B C}{A C \times B C}\right]$
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