A) \[-\frac{{{Q}^{2}}}{8\pi {{\varepsilon }_{0}}R}\]
B) \[-\frac{{{Q}^{2}}}{2\sqrt{2}\pi {{\varepsilon }_{0}}R}\]
C) \[-\frac{{{Q}^{2}}}{16\pi {{\varepsilon }_{0}}R}\]
D) \[-\frac{{{Q}^{2}}}{2\sqrt{2}\pi {{\varepsilon }_{0}}R}\]
Correct Answer: C
Solution :
[c] The electric field in initial and final configuration is as shown below:You need to login to perform this action.
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