The dimensional formula of $$\frac{1}{2} \epsilon_0 E^2$$ ($$\epsilon_0$$ = permittivity of vacuum and E = electric field) is $$M^a L^b T^c$$. The value of $$2a - b + c$$ = __________.

The numerical factor $$\frac12$$ is dimensionless, so the dimensions are governed only by $$\epsilon_0 E^2$$.

Step 1 : Dimensional formula of the electric field $$E$$
An electric field is force per unit charge: $$E = \frac{F}{Q}\,.$$
Force $$F$$ has the dimensions $$M L T^{-2}$$, while charge $$Q$$ has the dimensions of current × time, i.e. $$I T$$.
Hence
$$[E] = \frac{M L T^{-2}}{I T} = M L T^{-3} I^{-1}$$.

Step 2 : Dimensional formula of vacuum permittivity $$\epsilon_0$$
From Coulomb’s law, $$F = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q_1 Q_2}{r^2}\,,$$ so
$$\epsilon_0 = \frac{Q^2}{F\,r^{2}}$$ (ignoring the constant $$4\pi$$).
Therefore
$$[\epsilon_0] = \frac{(I T)^2}{\bigl(M L T^{-2}\bigr)\,L^{2}} = M^{-1} L^{-3} T^{4} I^{2}\,.$$

Step 3 : Dimensional formula of $$\epsilon_0 E^2$$
Square of the electric field:
$$[E^2] = \bigl(M L T^{-3} I^{-1}\bigr)^{2} = M^{2} L^{2} T^{-6} I^{-2}\,.$$
Multiply with $$[\epsilon_0]$$:
$$[\epsilon_0 E^2] = \bigl(M^{-1} L^{-3} T^{4} I^{2}\bigr) \bigl(M^{2} L^{2} T^{-6} I^{-2}\bigr) = M^{1} L^{-1} T^{-2}\,.$$

Thus $$\frac12 \epsilon_0 E^2$$ has the dimensional formula $$M^{a} L^{b} T^{c}$$ with
$$a = 1,\; b = -1,\; c = -2\,.$$

Step 4 : Required combination
$$2a - b + c = 2(1) - (-1) + (-2) = 2 + 1 - 2 = 1\,.$$

Hence the value of $$2a - b + c$$ is $$1$$.
Option B which is: $$1$$