A) zero
B) \[\frac{Qqd}{2\pi {{\in }_{0}}{{\ell }^{3}}}\]
C) \[\frac{Qqd}{\pi {{\in }_{0}}{{\ell }^{3}}}\]
D) \[\frac{Qqd}{4\pi {{\in }_{0}}{{\ell }^{3}}}\]
Correct Answer: B
Solution :
The two plates acts as a dipole \[\therefore \] Force on charge q; \[F=Eq\] \[\Rightarrow \] \[\left( \frac{2kQd}{{{\ell }^{3}}} \right)\cdot q=\frac{Qqd}{2\pi {{\varepsilon }_{0}}{{\ell }^{3}}}.\]You need to login to perform this action.
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