A) capacitance
B) electric field
C) permittivity
D) electric potential
Correct Answer: A
Solution :
As given \[[a]=[M],\,[b]=[L],\,[c]=[T]\]and \[[d]=[A]\] Therefore, \[\frac{a{{b}^{2}}}{{{c}^{2}}d}=\frac{[M][{{L}^{2}}]}{[{{T}^{3}}][A]}=\frac{[M{{L}^{2}}{{T}^{-2}}]}{[AT]}\]\[=\frac{work}{\text{charge}}=\] electric potentialYou need to login to perform this action.
You will be redirected in
3 sec