Question #0ed52

Jan 6, 2017

for part (a) charge $q$ on the capacitor is already maximum charge it can hold.
For part (b) potential difference $V$ across its plates is already at its maximum rated voltage.
We know that Capacitance $C$ is defined in terms of stored charge in a capacitor as
$C \equiv \frac{Q}{V}$
where $Q =$ magnitude of charge stored on each plate of the capacitor, and $V =$ voltage applied to the plates. Its SI units are Farad $F$
(a) When charge is doubled. Assuming that there is no change in the applied voltage, as shown from the equation above, capacitance is doubled; $\times 2$.
(b) When the potential difference $V$ across plates it is tripled. Assuming that there is no change in the stored charge, from the equation above, capacitance becomes one third; $\times \frac{1}{3}$.