Question

A parallel-plate capacitor is charged by a 15.0 V battery, then the battery is removed.

A)What is the potential difference between the plates after the battery is disconnected?

B)What is the potential difference between the plates after a sheet of Teflon is inserted between them?

Answer 1

ONCE THE CAPACITOR IS CHARGED TO A CERTAIN POTENTIAL, THE POTENTIAL ONLY DECREASES WHEN CERTAIN ENERGY IS DRAINED FROM IT e.g. IT IS CONNECTED TO A RESISTANCE IN A CLOSED CIRCUIT.

IN THE PRESENT CASE, AFTER CHARGING THE CAPACITOR, THE BATTERY IS DISCONNECTED. THEREFORE NOW THE CAPACITOR IS NOT CONNECTED TO ANY EXTERNAL LOAD OR IT CAN BE SAID THAT WE ARE HAVING AN OPEN CIRCUIT.

THEREFORE IN CASE (A) THE POTENTIAL DIFFERENCE BETWEEN PLATES AFTER THE BATTERY IS DISCONNECTED WILL REMAIN UNALTERED OR IT WILL BE AGAIN 15.0 Volts

LET US TAKE CASE (B) WHEREIN, TEFLON SHEET IS INSERTED BETWEEN THE PLATES. THE CAPACITANCE C' IN THIS CASE WILL BE C'= , WHERE C IS THE ORIGINAL CAPACITANCE AND IS THE DI-ELECTRIC CONSTANT OF TEFLON.

SINCE CHARGE STORED BEFORE AND AFTER INSERTION OF TEFLON WILL REMAIN SAME, HENCE Q=CV=C'V', WHERE C AND V ARE CAPACITANCE AND VOLTAGE BEFORE INSERTION OF TEFLON RESPECTIVELY AND C' AND V' ARE CAPACITANCE AND VOLTAGE AFTER INSERTION OF TEFLON RESPECTIVELY.

THEREFORE Q=C*V= ...................(1)

FROM EQUATION 1 WE GET

THEREFORE WE INFER THAT BY INSERTION OF TEFLON OF DIELECTRIC , THE VOLTAGE ACROSS PLATE OF CAPACITOR IS REDUCED TO