Q 11-01-062JEE MainJEE Main 2026 (2 Apr, Shift 1)Easy
The dimensional formula of $\frac{1}{2}\epsilon_0E^2$ ($\epsilon_0$ = permittivity of vacuum and $E$ = electric field) is $M^aL^bT^c$.
The value of $2a-b+c=$ ______.
Answer: (B) $1$
$\frac{1}{2}\epsilon_0E^2$ is the energy stored per unit volume of an electric field, so it has the dimensions of energy/volume:
$$\left[\frac{\text{energy}}{\text{volume}}\right]=\frac{ML^2T^{-2}}{L^3}=ML^{-1}T^{-2}$$
So $a=1$, $b=-1$, $c=-2$ and $2a-b+c=2+1-2=1$.
Solution by Sreeraj P, M.Sc Physics