What happens when you double the separation between a parallel capacitor?
The full question states, A parallel plate capacitor is connected to a battery. If we double the plate separation,
1. the capacitance is doubled
2. None of these
3. the charge on each plate is halved
4. the potential difference is halved
5. the Electric field is doubled
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The applicable electrical relations are:
C = εA/d, Q = CV, V = Q/C, E = V/d
if d’ = 2d
then C’ = εA/d’ = εA/2d = ½ C, => capacitance halved.
Q’ = C’V = ½ CV, then Q’ = ½ Q => charge halved
V potential difference unchanged.
E’ = V/d’ = V/2d = ½ V/d = ½ E, => electric field halved.
The correct choice is 3; the charge on each plate is halved.
B.
Doing so will cause the entire universe to collapse into itself, which will then expand to create a new, better universe because you wouldn’t be there to ruin it with such ignorant questions.
Capacitance is proportional to the area of the capacitor surface and inversely proportional to the distance between the plates:
C ~= a/d
thus if you double distance you’ll be dividing the capacitance by two.
Charge equals capacitance times voltage, so assuming a constant voltage, the charge on each plate will be halved.