Question #0ed52

1 Answer
A08
Jan 6, 2017

The answer assumes that
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-=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; #xx2#.

(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; #xx1/3#.

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