Question

Two charges $+6\mu C$ and $-4\mu C$ are placed $15cm$ apart as shown. At what distances from $A$ to its right, the electrostatic potential is zero?

• A. $4,9,60$
• B. $9,15,45$
• C. $20,30,40$
• D. $9,45$, infinity

Answer 1

Answer: (D)

Let the potential be zero at point $P$ at a distance $x$ from the charge $+6\times {10}^{-6}C$ at $A$ as shown in above figure. Potential at $P$
$V=\frac{1}{4\pi {\epsilon }_0}[\frac{6\times {10}^{-6}}{x}-\frac{(-4\times {10}^{-6})}{15-x}]$
$0=\frac{1}{4\pi {\epsilon }_0}[\frac{6\times {10}^{-6}}{x}-\frac{4\times {10}^{-6}}{15-x}]$
$0=\frac{6\times {10}^{-6}}{x}-\frac{4\times {10}^{-6}}{15-x}$
$\frac{6\times {10}^{-6}}{x}=\frac{4\times {10}^{-6}}{15-x}$
$\Rightarrow 6(15-x)=4x$
$\Rightarrow 90-6x=4x$
$\Rightarrow 10x=90$
$\Rightarrow x=\frac{90}{10}=9cm$
The other possibility is that point of zero potential $P$ may lie on $AB$ produced at a distance $x$ from the charge $+6\times {10}^{-6}C$ at $A$ as shown in the figure.
Potential at $P$
$V=\frac{1}{4\pi {\epsilon }_0}[\frac{6\times {10}^{-6}}{x}+\frac{(-4\times {10}^{-6})}{(15-x)}]$
$0=\frac{1}{4\pi {\epsilon }_0}[\frac{6\times {10}^{-6}}{x}-\frac{4\times {10}^{-6}}{(x-15)}]$
$0=\frac{6\times {10}^{-6}}{x}-\frac{4\times {10}^{-6}}{(x-15)}$
$\Rightarrow \frac{6\times {10}^{-6}}{x}=\frac{4\times {10}^{-6}}{x-15}$
$\Rightarrow \frac{6}{x}=\frac{4}{x-15}$
$\Rightarrow 6x-90=4x$
$\Rightarrow 2x=90$
$\Rightarrow x=\frac{90}{2}=45cm$