Dimensional Analysis of Energy Density of Electric Field

To find the dimensional formula of the expression $\frac{1}{2} \epsilon_0 E^2$, we recognize that this quantity represents the energy density ($u$) of an electric field, which is defined as energy per unit volume:

$u = \frac{\text{Energy}}{\text{Volume}}$

1. Dimensional formula of Energy: $\text{Energy} = \text{Force} \times \text{Displacement}$ $[\text{Energy}] = [M L T^{-2}] \times [L] = M^1 L^2 T^{-2}$

2. Dimensional formula of Volume: $[\text{Volume}] = L^3$

3. Dimensional formula of Energy Density ($\frac{1}{2} \epsilon_0 E^2$): $\left[\frac{1}{2} \epsilon_0 E^2\right] = \frac{M^1 L^2 T^{-2}}{L^3} = M^1 L^{-1} T^{-2}$

Comparing this with the given dimensional formula $M^a L^b T^c$, we get: $a = 1$ $b = -1$ $c = -2$

Now, substituting these values into the required expression $2a - b + c$: $2a - b + c = 2(1) - (-1) + (-2)$ $2a - b + c = 2 + 1 - 2 = 1$

Thus, the value of $2a - b + c$ is $1$.

Correct Option: B