Question

Two charges $3\times {10}^{-8}C$ and $-2\times {10}^{-8}$ are located $15cm$ apart. At what point on the line joining the two charges is the electric potential is zero?

• A. $9cm$ from $-2\times {10}^{-8}C$
• B. $10cm$ from $-2\times {10}^{-8}C$
• C. $9cm$ from $3\times {10}^{-8}C$
• D. $10cm$ from $-3\times {10}^{-8}C$

Answer 1

The correct option is C $9cm$ from $3\times {10}^{-8}C$

$V_T=0$

$\Rightarrow \frac{k(3\times {10}^{-8})}{x}+\frac{k(-2\times {10}^{-8})}{15-x}=0$

$x=9cm$