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 {{\varepsilon }_{0}}}\left[ \frac{6\times {{10}^{-6}}}{x}-\frac{(-4\times {{10}^{-6}})}{15-x} \right] $ $ 0=\frac{1}{4\pi {{\varepsilon }_{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 {{\varepsilon }_{0}}}\left[ \frac{6\times {{10}^{-6}}}{x}+\frac{(-4\times {{10}^{-6}})}{(15-x)} \right] $ $ 0=\frac{1}{4\pi {{\varepsilon }_{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 $