A parallel-plate capacitor is connected to a battery and stores 3.6 nC of charge

Question

A parallel-plate capacitor is connected to a battery and stores 3.6 nC 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. Does the capacitor's charge increase or decrease? The capacitor's charge increases. The capacitor's charge decreases. The capacitor's charge remains the same. It's impossible to determine. By how much does the charge change? Express your answer with the appropriate units.

Explanation / Answer

let the battery has emf of = V volts

so charge stored

Q1 = CV

V = Q1/C

now after dielectric capacitance will change

C2 = kC

so new charge Q2 = C2 V

Q2 = (kC) * Q1/C

Q2 = kQ1

Q2 = 2.1 * 3.6 nC             { we know that for teflon dielectric k = 2.1}

Q2 = 7.56 nC

1) the capacitor charge will increase

2) change in charge = Q2-Q1

= (7.56 - 3.6 ) nC

= 3.96 C

answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---