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.