Question

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

• A. $\text{4},\text{ 9},\text{ 6}0$
• B. $\text{9},\text{ 15},\text{45}$
• C. $\text{2}0,\text{ 3}0,\text{ 4}0$
• D. $\text{9}.\text{ 45},$ 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 $V=\frac{1}{4\pi {\epsilon }_0}\left[\frac{6\times {10}^{-6}}{x}-\frac{(-4\times {10}^{-6})}{15-x}\right]$ $0=\frac{1}{4\pi {\epsilon }_0}\left[\frac{6\times {10}^{-6}}{x}-\frac{4\times {10}^{-6}}{15-x}\right]$ $0=\frac{6\times {10}^{-6}}{x}-\frac{4\times {10}^{-6}}{15-x}$ $\frac{6\times {10}^{-6}}{x}=\frac{4\times {10}^{-6}}{15-x}$ $\Rightarrow$ $6(15-x)=4x$ $\Rightarrow$ $90-6x=4x$ $\Rightarrow$ $10x=90$ $\Rightarrow$ $x=\frac{90}{10}=9cm$ 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 m the figure.
Potential at P $V=\frac{1}{4\pi {\epsilon }_0}\left[\frac{6\times {10}^{-6}}{x}+\frac{(-4\times {10}^{-6})}{(15-x)}\right]$ $0=\frac{1}{4\pi {\epsilon }_0}\left[\frac{6\times {10}^{-6}}{x}-\frac{4\times {10}^{-6}}{(x-15)}\right]$ $0=\frac{6\times {10}^{-6}}{x}-\frac{4\times {10}^{-6}}{(x-15)}$ $\Rightarrow$ $0=\frac{6\times {10}^{-6}}{x}=\frac{4\times {10}^{-6}}{(x-15)}$ $\Rightarrow$ $\frac{6}{x}=\frac{4}{x-15}$ $\Rightarrow$ $6x-90=4x$ $\Rightarrow$ $2x=90$ $\Rightarrow$ $x=\frac{90}{2}=45cm$