Question

A parallel-plate capacitor is connected to a battery and stores 4.0 nCnC of charge. Then, while the battery remains connected, a sheet of Teflon is inserted between the plates. For the dielectric constant, use the value from Table 21.3. By how much does the charge change?

Answer 1

Given that initial charge on capacitor, Q0 = 4.0 nC

Now charge is given by:

Q0 = C0*V0

Since capacitor remain coonected to the battery while insertion of dielectric, So potential across capacitor remains constant. Now when Teflon sheet is inserted, then capacitance of capacitor will be:

C1 = k*C0

k = dielecteic constant of Teflon = 2.1

So C1 = 2.1*C0

Now new charge on capacitor will be:

Q1 = C1*V1

Since V1 = V0 & C1 = k*C0 = 2.1*C0

So,

Q1 = 2.1*C0*V0

Q1 = 2.1*Q0

Now change in value of charge will be:

dQ = Q1 - Q0 = 2.1*Q0 - Q0

dQ = 1.1*Q0

dQ = 1.1*4.0

dQ = 4.4 nC = change in charge