Question

The dimensional formula of capacitance is:

• A. $[M^{-1}L^{-2}T^4A^2]$
• B. $[ML^2T^{-2}A^{-2}]$
• C. $[M^{-1}L^{-1}T^3A]$
• D. $[MLT^{-3}A^{-1}]$

Hint:

Capacitance is highly storage-effective, meaning it has large positive powers for Time ($T^4$) and Current ($A^2$), while Mass ($M^{-1}$) and Length ($L^{-2}$) are negative. Memorizing the structural signature $[-1, -2, 4, 2]$ helps identify option (A) instantly.

Answer 1

The Correct Option is A

Concept: The capacitance ($C$) of a conductor is defined as the ratio of the magnitude of the electric charge ($Q$) on it to its electric potential ($V$): \[ C = \frac{Q}{V} \] By evaluating the dimensional formulas of charge and electric potential back to fundamental SI units (Mass $[M]$, Length $[L]$, Time $[T]$, and Electric Current $[A]$), we can derive the dimensions of capacitance.

Step 1: Find the dimensional formulas for Charge ($Q$) and Work ($W$).

• Electric charge is current multiplied by time ($Q = I \cdot t$): \[ [Q] = [A^1 T^1] \]
• Mechanical Work or Energy is force times distance ($W = F \cdot s = m \cdot a \cdot s$): \[ [W] = [M^1 L^2 T^{-2}] \]

Step 2: Derive the dimensional formula for Electric Potential ($V$).
Electric potential is defined as the work done per unit charge ($V = \frac{W}{Q}$): \[ [V] = \frac{[M^1 L^2 T^{-2}]}{[A^1 T^1]} = [M^1 L^2 T^{-3} A^{-1}] \]

Step 3: Calculate the dimensions of Capacitance ($C$).
Now substitute the dimensional formulas of $Q$ and $V$ back into our baseline capacitance definition ($C = \frac{Q}{V}$): \[ [C] = \frac{[A^1 T^1]}{[M^1 L^2 T^{-3} A^{-1}]} \] Using exponent rules to shift all base parameters to the numerator: \[ [C] = [M^{-1} L^{-2} T^{1 - (-3)} A^{1 - (-1)}] = [M^{-1} L^{-2} T^4 A^2] \]