Which one of the following is the correct dimensional formula for the capacitance in F ? M, L, T and C stand for unit of mass, length, time and charge,

The dimensional formula for capacitance in farads (F) can be derived using the definition of capacitance. Capacitance $$C$$ is given by the formula:

$$ C = \frac{Q}{V} $$

where $$Q$$ is the charge and $$V$$ is the potential difference.

Charge $$Q$$ has the dimension $$[C]$$, so:

$$ [Q] = [C] $$

Potential difference $$V$$ is defined as work done per unit charge:

$$ V = \frac{W}{Q} $$

where $$W$$ is work done. Work done $$W$$ is force multiplied by displacement:

$$ W = F \cdot d $$

Force $$F$$ is mass times acceleration:

$$ F = m \cdot a $$

Acceleration $$a$$ has dimension $$[LT^{-2}]$$, so force $$F$$ has dimension:

$$ [F] = [M] \cdot [LT^{-2}] = [MLT^{-2}] $$

Displacement $$d$$ has dimension $$[L]$$, so work done $$W$$ has dimension:

$$ [W] = [F] \cdot [d] = [MLT^{-2}] \cdot [L] = [ML^2T^{-2}] $$

Therefore, potential difference $$V$$ has dimension:

$$ [V] = \frac{[W]}{[Q]} = \frac{[ML^2T^{-2}]}{[C]} = [ML^2T^{-2}C^{-1}] $$

Now, capacitance $$C$$ is:

$$ [C] = \frac{[Q]}{[V]} = \frac{[C]}{[ML^2T^{-2}C^{-1}]} $$

Simplifying the expression:

$$ [C] = [C] \times \frac{1}{[ML^2T^{-2}C^{-1}]} = [C] \times [M^{-1}L^{-2}T^{2}C] = [C^{1} \cdot C^{1}] \times [M^{-1}L^{-2}T^{2}] = [C^{2}M^{-1}L^{-2}T^{2}] $$

Thus, the dimensional formula for capacitance is $$[C^{2}M^{-1}L^{-2}T^{2}]$$.

Comparing with the options:

• A. $$[F]=[C^{2}M^{-1}L^{-2}T^{2}]$$
• B. $$[F]=[C^{2}M^{-2}L^{2}T^{2}]$$
• C. $$[F]=[CM^{-2}L^{-2}T^{-2}]$$
• D. $$[F]=[CM^{-1}L^{-2}T^{2}]$$

Option A matches the derived dimensional formula.

Therefore, the correct answer is A.