Question

Farad is not equivalent to $(q=$ coulomb, $V=$ volt and $J=$ joule $)$

• A. $\frac{q}{V}$
• B. $q{V}^2$
• C. $\frac{q^2}{J}$
• D. $\frac{J}{V^2}$

Answer 1

Answer: (B)

Capacitance $C=\frac{\text{ Charge }}{\text{ Potential }}=\frac{q}{V}$
Also potential $=\frac{\text{ Work }}{\text{ Charge }}$
$\left(\because v=\frac{w}{q}\right)$
$\therefore C=\frac{q^2}{J}$ as well as $C=\frac{J}{V^2}$.
Thus, (a), (c), (d) are equivalent to farad but (b) equivalent to Farad. but (b) is not equivalent to Farad.