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The parallel plate capacitor can only store a finite amount of energy before dielectric breakdown occurs. It can be defined as: When two parallel plates are connected across a battery, the plates are charged and an electric field is established between them, and this setup is known as the parallel plate capacitor.

Given We are given the capacitance of parallel-plate capacitor then disconnected after fully charged_

Required

We are asked to calculate (a) The voltage Vab between the two plates _

(b) (i) The voltage Vab when the distance d is doubled

(ii) The voltage Vab when the radius r is doubled and the separated distance is constant

(a) The capacitor is charged with the battery by 120 V, this means the potential difference between the two parallel plates after fully charging i

Therefore the Voltmeter reading is 12V

(b) (i) The capacitance is related to the separated distance by equation 242 in the form

As shown by equation (l)

1 So when the distance is doubled this means the capacitance is halved. If the charge is constant, then the voltage will be doubled as the capacitance is halved where equation 24]

shows that, the capacitance is inversely proportional to the voltage

So as the capacitance is halved the voltage is doubled

Therefore voltmeter reading if plate separation is doubled is 24V

(ii) As shown by equation (1) the capacitance depends directly on the area A, where A equals 4arr2_ Therefore,

So, as the radius is doubled, the capacitance will increase four times when the separated distance is unchanged and

As shown by equation (2), the capacitance is inversely proportional to the potential difference Vab_ So as the capacitance increases four times, therefore the voltage will decrease by four times and becomes

Therefore if the radius of each plate were doubled but their separation was unchanged

is 3V