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 \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{6\times {{10}^{-6}}}{x}+\frac{(-4\times {{10}^{-6}})}{(15-x)} \right]\] \[0=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{6\times {{10}^{-6}}}{x}-\frac{4\times {{10}^{-6}}}{(x-15)} \right]\] \[0=\frac{6\times {{10}^{-6}}}{x}-\frac{4\times {{10}^{-6}}}{(x-15)}\] \[\Rightarrow \] \[0=\frac{6\times {{10}^{-6}}}{x}=\frac{4\times {{10}^{-6}}}{(x-15)}\] \[\Rightarrow \] \[\frac{6}{x}=\frac{4}{x-15}\] \[\Rightarrow \] \[6x-90=4x\] \[\Rightarrow \] \[2x=90\] \[\Rightarrow \] \[x=\frac{90}{2}=45cm\]You need to login to perform this action.
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