JEE PYQ: Units & Measurements - Question ID a1aff95f8cda (JEE Main 2026)

Core Formula & Concept:

In electromagnetism, the fundamental relation connecting electric field $E$, free-space permittivity $\epsilon_0$, and current density $J$ is given by Maxwell’s displacement-current term: $J_{\text{disp}} = \epsilon_0 \frac{\partial E}{\partial t}.$ This equation tells us that the quantity $\epsilon_0 E/t$ has the physical meaning of a current density (current per unit area). Consequently, its SI unit must be the same as that of current density, namely ampere per square metre ($\mathrm{A/m^2}$).

Step-by-Step Derivation:

Step 1: Write down the dimensions of each quantity.

• Free-space permittivity $\epsilon_0$ has the SI unit farad per metre ($\mathrm{F/m}$).
• Electric field $E$ has the SI unit volt per metre ($\mathrm{V/m}$).
• Time $t$ has the SI unit second ($\mathrm{s}$).

Step 2: Express each unit in terms of base SI units.

• $\mathrm{F} = \mathrm{C/V} = \mathrm{C^2/(N \cdot m)} = \mathrm{C^2 \cdot s^2/(kg \cdot m^3)}$.
• Therefore $\epsilon_0$ in base units is $\mathrm{C^2 \cdot s^2/(kg \cdot m^3) \cdot m^{-1}} = \mathrm{C^2 \cdot s^2/(kg \cdot m^2)}$.
• $E$ in base units is $\mathrm{V/m} = \mathrm{(J/C)/m} = \mathrm{(N \cdot m/C)/m} = \mathrm{N/C} = \mathrm{(kg \cdot m/s^2)/C}$.
• $t$ is simply $\mathrm{s}$.

Step 3: Form the combination $\epsilon_0 E / t$ and simplify.

$$\frac{\epsilon_0 E}{t} = \frac{\bigl[\mathrm{C^2 \cdot s^2/(kg \cdot m^2)}\bigr] \cdot \bigl[\mathrm{(kg \cdot m/s^2)/C}\bigr]}{\mathrm{s}} = \frac{\mathrm{C^2 \cdot s^2 \cdot kg \cdot m \cdot s^{-2} \cdot C^{-1}}}{\mathrm{kg \cdot m^2 \cdot s}} = \frac{\mathrm{C}}{ \mathrm{m^2 \cdot s} }.$$

Since $1\,\mathrm{C/s} = 1\,\mathrm{A}$, the unit becomes $\mathrm{A/m^2}$.

Step 4: Match with the given options.

The derived unit $\mathrm{A/m^2}$ corresponds exactly to option C.

Common Traps & Exam Tip:

Many students mistakenly treat $\epsilon_0$ as dimensionless or confuse its unit with that of capacitance. A frequent error is to cancel $\mathrm{F}$ and $\mathrm{V}$ incorrectly, leading to options like $\mathrm{A/m}$ (option B). Always reduce every quantity to base SI units (kg, m, s, A) before combining them to avoid such pitfalls.