Question

Two charges $+6\mu C$ and $-4\mu C$ are placed apart as shown. At what distances from $A$ to its right, the electrostatic potential is zero (distances in cm)

• A. 4, 9 , and 60
• B. 9, 45, infinity
• C. 20, 45, infinity
• D. 9, 15, 45

Answer 1

Answer: (B)

$V_1+V_2=0$
$\frac{Q_1}{4\pi {\epsilon }_0{r}_1}+\frac{Q_2}{4\pi {\epsilon }_0{r}_2}=0$
$\frac{1}{4\pi {\epsilon }_0}\left[\frac{6\times 10^{-6}}{x}+\frac{-4\times 10^{-6}}{\left(15-x\right)}\right]=0$
$9\times 10^9\left[\frac{6\times 10^{-6}}{x}+\frac{\left(-4\times 10^{-6}\right)}{\left(15-x\right)}\right]=0$
$\frac{6\times 10^{-6}}{x}=\frac{4\times 10^{-6}}{\left(15-x\right)}$
$6\left(15-x\right)=4x$
$90-6x=4x$
$10x=90$
$x=9cm$
Again at $45cm$
Potential $V=\frac{6\times 10^{-6}}{45}+\frac{(-4\times 10^{-6})}{30}$
$=0.13\times 6^{-6}-0.13\times 10^{-6}$
$=0$
At infinity
Potential $V=\frac{6\times 10^{-6}}{\infty }+\frac{4\times 10^{-6}}{\infty }=0$