JEE PYQ: Units & Measurements - Question ID 5ace8df8781f (JEE Main 2013)

The permittivity of vacuum ($\varepsilon_0$) is a fundamental constant in electromagnetism that appears in Coulomb’s law and Gauss’s law for electric fields. Its dimensional formula can be derived using the force between two point charges.

The key formula we use is Coulomb’s law: $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$ where

• $F$ is the electrostatic force (dimensions: $[M L T^{-2}]$),
• $q_1, q_2$ are electric charges (dimensions: $[A T]$),
• $r$ is the distance between the charges (dimensions: $[L]$).

Our goal is to isolate $\varepsilon_0$ and express its dimensions in terms of mass ($M$), length ($L$), time ($T$), and electric current ($A$).

Step 1: Start from Coulomb’s law and solve for $\varepsilon_0$: $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} \implies \varepsilon_0 = \frac{1}{4 \pi} \frac{q_1 q_2}{F r^2}$ Since $4 \pi$ is a dimensionless constant, it does not affect the dimensional formula. Thus, we focus on: $\varepsilon_0 \propto \frac{q_1 q_2}{F r^2}$

Step 2: Substitute the dimensions of each quantity:

• Charge ($q$): $[A T]$
• Force ($F$): $[M L T^{-2}]$
• Distance ($r$): $[L]$

Step 3: Simplify the expression: $[\varepsilon_0] = [A^2 T^2] \cdot [M^{-1} L^{-3} T^{2}] = [M^{-1} L^{-3} T^{4} A^{2}]$

Step 4: Compare with the given options:

• Option A: $[M^{-1} L^{-3} T^{2} A]$ → Incorrect (missing $T^2$ and $A$ exponent is wrong).
• Option B: $[M^{-1} L^{-3} T^{4} A^{2}]$ → Matches our derived formula.
• Option C: $[M^{1} L^{2} T^{1} A^{2}]$ → Incorrect (signs and exponents are wrong).
• Option D: $[M^{1} L^{2} T^{1} A]$ → Incorrect (signs and exponents are wrong).

Trap 1: Forgetting that charge ($q$) has dimensions $[A T]$. Some students mistakenly assume charge is dimensionless or has dimensions of current alone ($[A]$), leading to incorrect exponents for $A$ and $T$.

Trap 2: Misapplying the force formula. Students sometimes confuse electrostatic force with magnetic force or gravitational force, leading to incorrect dimensional substitutions.

Trap 3: Ignoring the $4 \pi$ factor. While it is dimensionless, some students unnecessarily complicate the derivation by trying to account for it dimensionally.

Exam Tip: Always cross-verify dimensions by plugging them back into the original formula. For example, substitute $[\varepsilon_0] = [M^{-1} L^{-3} T^{4} A^{2}]$ into Coulomb’s law and confirm that both sides have the same dimensions ($[M L T^{-2}]$).