A) electric field
B) capacity
C) potential
D) permittivity
Correct Answer: C
Solution :
We know that Potential \[V=\frac{energy\,\left( Q \right)}{charge\text{ }\left( q \right)}\] Dimensions of energy \[(Q)=[M{{L}^{2}}{{T}^{-2}}]\] Dimensions of charge (q) = [IT] \[\therefore \] Dimensions of potential \[V=\frac{[M{{L}^{2}}{{T}^{-2}}]}{[IT]}\] Dimensions of potential \[V=[M{{L}^{2}}{{T}^{-3}}{{I}^{-1}}]\]You need to login to perform this action.
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