Question #0ed52
1 Answer
The answer assumes that
for part (a) charge
For part (b) potential difference
We know that Capacitance
#C-=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;
(b) When the potential difference