JEE PYQ: Units & Measurements - Question ID 36a18d251d62 (JEE Main 2024)

Core Formula & Concept:

In electrostatics, the energy density stored in an electric field $E$ is given by the expression: $u = \frac{1}{2} \epsilon_{\mathrm{o}} E^2$ where

• $\epsilon_{\mathrm{o}}$ is the permittivity of free space, and
• $E$ is the magnitude of the electric field.

Energy density $u$ has dimensions of energy per unit volume, i.e. $[u] = \mathrm{M\,L^{-1}\,T^{-2}}$. Since the factor $\tfrac12$ is dimensionless, the combination $\epsilon_{\mathrm{o}} E^2$ must carry the same dimensions as $u$.

Step-by-Step Derivation:

1. Write the dimensions of permittivity $\epsilon_{\mathrm{o}}$. From Coulomb’s law, force $F = \dfrac{1}{4\pi\epsilon_{\mathrm{o}}}\,\dfrac{q_1 q_2}{r^2}$. Rearranging gives $\epsilon_{\mathrm{o}} = \frac{q_1 q_2}{4\pi F\,r^2}.$ Charge $q$ has dimensions $\mathrm{A\,T}$ (ampere-second), force $F$ is $\mathrm{M\,L\,T^{-2}}$, and distance $r$ is $\mathrm{L}$. Therefore $[\epsilon_{\mathrm{o}}] = \frac{(\mathrm{A\,T})^2}{\mathrm{M\,L\,T^{-2}}\cdot \mathrm{L^2}} = \mathrm{M^{-1}\,L^{-3}\,T^4\,A^2}.$ 2. Write the dimensions of electric field $E$. By definition $E = F/q$, so $[E] = \frac{\mathrm{M\,L\,T^{-2}}}{\mathrm{A\,T}} = \mathrm{M\,L\,T^{-3}\,A^{-1}}.$ Squaring gives $[E^2] = \bigl(\mathrm{M\,L\,T^{-3}\,A^{-1}}\bigr)^2 = \mathrm{M^2\,L^2\,T^{-6}\,A^{-2}}.$ 3. Multiply the dimensions of $\epsilon_{\mathrm{o}}$ and $E^2$: $[\epsilon_{\mathrm{o}} E^2] = [\epsilon_{\mathrm{o}}] \cdot [E^2] = \bigl(\mathrm{M^{-1}\,L^{-3}\,T^4\,A^2}\bigr) \cdot \bigl(\mathrm{M^2\,L^2\,T^{-6}\,A^{-2}}\bigr).$ Combine exponents term by term:

• Mass: $-1 + 2 = +1$
• Length: $-3 + 2 = -1$
• Time: $+4 - 6 = -2$
• Ampere: $+2 - 2 = 0$

Hence $[\epsilon_{\mathrm{o}} E^2] = \mathrm{M^1\,L^{-1}\,T^{-2}\,A^0} = \mathrm{M\,L^{-1}\,T^{-2}}.$ 4. Compare with the given options. Option A is $\mathrm{M\,L^{-1}\,T^{-2}}$, which matches exactly.

Common Traps & Exam Tip:

• Forgetting to square $E$: Students sometimes compute $[\epsilon_{\mathrm{o}} E]$ instead of $[\epsilon_{\mathrm{o}} E^2]$, leading to an incorrect dimension.
• Miscounting exponents: When multiplying dimensions, it is easy to add exponents incorrectly. Always write out each base (M, L, T, A) separately.
• Confusing energy density with energy: Energy has dimensions $\mathrm{M\,L^2\,T^{-2}}$, but energy density is energy per unit volume, $\mathrm{M\,L^{-1}\,T^{-2}}$.

Remember that $\epsilon_{\mathrm{o}} E^2$ represents an energy density, so its dimensions must be $\mathrm{M\,L^{-1}\,T^{-2}}$. This immediately points to option A.