JEE PYQ: Units & Measurements - Question ID b700fa7df1c4 (JEE Main 2026)

Core Formula & Concept:

The problem involves determining the dimensional formula of the expression $\frac{1}{2} \epsilon_0 E^2$, where:

• $\epsilon_0$ is the permittivity of free space (vacuum).
• $E$ is the electric field.

This expression represents the energy density stored in an electric field in vacuum. To find its dimensions, we must recall the dimensional formulas of $\epsilon_0$ and $E$.

Key dimensional formulas:

• Force: $F = ma \implies [F] = M L T^{-2}$
• Electric field $E$ is defined as force per unit charge: $E = \frac{F}{q} \implies [E] = \frac{[F]}{[q]}$
• Charge $q$ has dimensions $[q] = I T$ (current × time).
• Permittivity $\epsilon_0$ appears in Coulomb’s law: $F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$. From this, we derive $[\epsilon_0] = \frac{[q]^2}{[F][L]^2} = \frac{(I T)^2}{(M L T^{-2})(L^2)} = M^{-1} L^{-3} T^4 I^2$.

Step-by-Step Derivation:

Step 1: Write the expression and identify dimensions
We are to find the dimensions of $\frac{1}{2} \epsilon_0 E^2$. Since the factor $\frac{1}{2}$ is dimensionless, we focus on $\epsilon_0 E^2$.

Step 2: Substitute dimensional formulas
From above:

• $[\epsilon_0] = M^{-1} L^{-3} T^4 I^2$
• $[E] = \frac{[F]}{[q]} = \frac{M L T^{-2}}{I T} = M L T^{-3} I^{-1}$

Thus, $[\epsilon_0 E^2] = [\epsilon_0] \cdot [E]^2 = (M^{-1} L^{-3} T^4 I^2) \cdot (M L T^{-3} I^{-1})^2$

Step 3: Expand $[E]^2$
$[E]^2 = (M L T^{-3} I^{-1})^2 = M^2 L^2 T^{-6} I^{-2}$

Step 4: Multiply dimensions
$[\epsilon_0 E^2] = (M^{-1} L^{-3} T^4 I^2) \cdot (M^2 L^2 T^{-6} I^{-2}) = M^{(-1+2)} L^{(-3+2)} T^{(4-6)} I^{(2-2)} = M^1 L^{-1} T^{-2} I^0$ Since $I^0 = 1$, the dimensions simplify to: $[\epsilon_0 E^2] = M^1 L^{-1} T^{-2}$

Step 5: Match with given form and find $a$, $b$, $c$
The problem states the dimensional formula is $M^a L^b T^c$. Comparing:

• $a = 1$
• $b = -1$
• $c = -2$

Step 6: Compute $2a - b + c$
$2a - b + c = 2(1) - (-1) + (-2) = 2 + 1 - 2 = 1$

Common Traps & Exam Tip:

Students often make the following mistakes:

• Ignoring current dimension $I$: Some forget that $\epsilon_0$ has dimensions involving current, leading to incorrect cancellation. Always include $I$ in dimensional analysis of electromagnetic quantities.
• Incorrect sign in exponents: Misapplying exponent rules (e.g., $(M L T^{-3})^2 = M^2 L^2 T^{-6}$, not $T^{-3}$) leads to wrong values of $a$, $b$, $c$.
• Forgetting $\frac{1}{2}$ is dimensionless: Some mistakenly try to assign dimensions to the numerical factor. Remember: pure numbers are dimensionless.

Always double-check each step of dimensional multiplication and ensure all exponents are correctly combined.

Final Answer: The value of $2a - b + c$ is 1, which corresponds to option B.