question_answer

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

Correct Answer: D

Solution :

  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 \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{6\times {{10}^{-6}}}{x}+\frac{(-4\times {{10}^{-6}})}{(15-x)} \right]\] \[0=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{6\times {{10}^{-6}}}{x}-\frac{4\times {{10}^{-6}}}{(x-15)} \right]\] \[0=\frac{6\times {{10}^{-6}}}{x}-\frac{4\times {{10}^{-6}}}{(x-15)}\] \[\Rightarrow \]               \[0=\frac{6\times {{10}^{-6}}}{x}=\frac{4\times {{10}^{-6}}}{(x-15)}\] \[\Rightarrow \]               \[\frac{6}{x}=\frac{4}{x-15}\] \[\Rightarrow \]               \[6x-90=4x\] \[\Rightarrow \]               \[2x=90\] \[\Rightarrow \]               \[x=\frac{90}{2}=45cm\]