A) 0
B) \[\frac{4q}{d}\]
C) \[\frac{4d}{q}\]
D) \[\frac{q}{4d}\]
Correct Answer: B
Solution :
Potential at centre due to all charges are \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{q}{d}+\frac{q}{d}+\frac{q}{d}+\frac{q}{d} \right]\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4q}{d}\]in SI units \[=\frac{4q}{d}\]in CGS unitsYou need to login to perform this action.
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