A) \[V-ER\theta \]
B) \[V-2ER\,\sin \,\frac{\theta }{2}\]
C) \[V+ER\theta \]
D) \[V+2ER\,\sin \,\frac{\theta }{2}\]
Correct Answer: A
Solution :
As, \[V=-\int{E\cdot dr}\] \[\Rightarrow \] \[{{V}_{B}}-{{V}_{A}}=-ER\,\theta \] \[{{V}_{B}}=V-ER\,\theta \]You need to login to perform this action.
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