S2. A parallel plate capacitor with circular plates of radius R. separated by a

Question

S2. A parallel plate capacitor with circular plates of radius R. separated by a distance d R, is charged using a battery Being careful not to discharge the plates, it is then disconnected from the battery. a) How would the charge on the plates, the voltage difference across tne capacitor and the electric field inside the capacitor be affected by increasing the plate separation to 20? [3 marks Changes? changes, by whar Brig explanarion Voltage across capacitor Charge on plates E-field strength between plates

Explanation / Answer

a)

For a capacitor, Q = C*V

where Q = charge on the capacitor

C = capacitance = e*A/d

where e = permittivity of free space <--- constant

A = area of plates

d= plate separation

V = voltage across the capacitor

Now, as the plate separation is made 2d, the charge remains same (due to conservation of charge).

So, the capacitance becomes half as 'd' becomes '2d' which is in the denominator of the expression of C

But for Q to remain same , V must get doubled so as to negate the effect of C getting halved

So, V is doubled <--------answer

b)

When the plates are moved to a separation of 2d, the charge remains same. This is due to the fact that charge needs to be conserved as there is no circuit for the charges to discharge.

So, no change <--------answer

c)

For the capacitor, V = E*d

So, E = V/d

where E = electric field

As, d is doubled, V is also doubled, so the Electric field remains same <-------answer

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