A) \[V=\frac{p\cos \theta }{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]
B) \[V=\frac{p\cos \theta }{4\pi {{\varepsilon }_{0}}r}\]
C) \[V=\frac{p\sin \theta }{4\pi {{\varepsilon }_{0}}r}\]
D) \[V=\frac{p\cos \theta }{2\pi {{\varepsilon }_{0}}{{r}^{2}}}\]
Correct Answer: A
Solution :
For the given situation, diagram can be drawn as follows As shown in figure component of dipole moment along the line OP will be \[p'=p\cos \theta \]. Hence electric potential at point P will be \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{p\cos \theta }{{{r}^{2}}}\]You need to login to perform this action.
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