question_answer

Two charges \[+6\,\,\mu C\] and \[-4\,\,\mu C\] are placed 15 cm 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

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                 \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{6\times {{10}^{-6}}}{x}-\frac{(-4\times {{10}^{-6}})}{15-x} \right]\]                 \[0=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{6\times {{10}^{-6}}}{x}-\frac{4\times {{10}^{-6}}}{15-x} \right]\]                 \[0=\frac{6\times {{10}^{-6}}}{x}-\frac{4\times {{10}^{-6}}}{15-x}\] \[\frac{6\times {{10}^{-6}}}{x}=\frac{4\times {{10}^{-6}}}{15-x}\] \[\Rightarrow \] \[6(15-x)=4x\] \[\Rightarrow \] \[90-6x=4x\] \[\Rightarrow \] \[10x=90\] \[\Rightarrow \] \[x=\frac{90}{10}=9\,cm\] 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                 \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{6\times {{10}^{-6}}}{x}+\frac{(-4\times {{10}^{-6}})}{(15-x)} \right]\]                 \[0=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{6\times {{10}^{-6}}}{x}-\frac{4\times {{10}^{-6}}}{(x-15)} \right]\]                 \[0=\frac{6\times {{10}^{-6}}}{x}-\frac{4\times {{10}^{-6}}}{(x-15)}\] \[\Rightarrow \] \[\frac{6\times {{10}^{-6}}}{x}=\frac{4\times {{10}^{-6}}}{x-15}\] \[\Rightarrow \] \[\frac{6}{x}=\frac{4}{x-15}\] \[\Rightarrow \] \[6x-90=4x\] \[\Rightarrow \] \[2x=90\] \[\Rightarrow \] \[x=\frac{90}{2}=45\,\,cm\]