An initially uncharged air-filled capacitor is connected to a 4.37-V charging so
ID: 1448825 • Letter: A
Question
An initially uncharged air-filled capacitor is connected to a 4.37-V charging source. As a result, 7.25 × 10-5 C of charge is transfered from one of the capacitor's plates to the other. Then, while the capacitor remains connected to the charging source, a sheet of dielectric material is inserted between its plates, completely filling the space. The dielectric constant of this substance is 7.55. Find the capacitor's potential difference and charge after the insertion. Potential difference after insertion of dielectric: Charge after insertion of dielectric:
Explanation / Answer
V = initial potential difference = 4.37 volts
Q = initial charge stored = 7.25 x 10-5 C
C = capacitance
using the formula
Q = CV
C = Q/V = ( 7.25 x 10-5 ) / (4.37) = 1.66 x 10-5 F
New capacitance after dielectric insertion is given as
C' = kC = 7.55 x 1.66 x 10-5 = 12.53 x 10-5 F
New charge stored
Q' = C'V = 12.53 x 10-5 x 4.37 = 54.76 x 10-5 C=