Question

To measure capacitance of a parallel plate capacitor it is first charged to a potential $V_0$ =1350 V. It is then connected in parallel to a capacitor of identical geometry but filled with dielectric of dielectric constant k. As a result the voltage of $1^{st}$ capacitor become 450 volt and charge on $2^{nd}$ capacitor become 18 mc. Choose the correct option(s).

• A. initial charge on $1^{st}$ capacitor is 27 mC
• B. dielectric constant k = 2
• C. charge on $1^{st}$ capacitor after connection is 18 mC
• D. heat loss is zero in this process.

Answer 1

Answer: (A), (B)

initial charge on $1^{st}$ capacitor is 27 mC, dielectric constant k = 2

Answer 2

Let the capacitance of the first parallel plate capacitor (air-filled) be $C_1 = C$. The initial potential of the first capacitor is $V_0 = 1350$ V. The initial charge on the first capacitor is $Q_{1,initial} = C_1 V_0 = C \times 1350$.

The second capacitor has identical geometry but is filled with a dielectric of dielectric constant $k$. Its capacitance is $C_2 = kC$. Initially, it is uncharged.

When the two capacitors are connected in parallel, they share a common final potential $V_f = 450$ V. The charge on the first capacitor after connection is $Q_{1,final} = C_1 V_f = C \times 450$. The charge on the second capacitor after connection is $Q_{2,final} = C_2 V_f = kC \times 450$. We are given that $Q_{2,final} = 18$ mC $= 18 \times 10^{-3}$ C. So, $kC \times 450 = 18 \times 10^{-3}$. $kC = \frac{18 \times 10^{-3}}{450} = 40 \times 10^{-6}$ F.

By conservation of charge, the total initial charge equals the total final charge. The initial total charge is $Q_{total,initial} = Q_{1,initial} + Q_{2,initial} = C V_0 + 0 = C \times 1350$. The final total charge is $Q_{total,final} = Q_{1,final} + Q_{2,final} = C V_f + kC V_f = (C + kC) V_f$. Equating the initial and final total charges: $C \times 1350 = (C + kC) \times 450$. Divide by 450: $C \times 3 = C + kC$. $3C = C + kC$. $2C = kC$. Since $C$ is the capacitance of a capacitor, $C \neq 0$. We can divide by $C$: $k = 2$.

Now we can find $C$ using the value of $kC$: $kC = 2C = 40 \times 10^{-6}$ F. $C = \frac{40 \times 10^{-6}}{2} = 20 \times 10^{-6}$ F $= 20$ µF.

Let's evaluate the given options:

• initial charge on $1^{st}$ capacitor is 27 mC. $Q_{1,initial} = C V_0 = (20 \times 10^{-6} \text{ F}) \times (1350 \text{ V}) = 27000 \times 10^{-6}$ C $= 27 \times 10^{-3}$ C $= 27$ mC. This option is correct.

• dielectric constant k = 2. We calculated $k=2$. This option is correct.

• charge on $1^{st}$ capacitor after connection is 18 mC. $Q_{1,final} = C V_f = (20 \times 10^{-6} \text{ F}) \times (450 \text{ V}) = 9000 \times 10^{-6}$ C $= 9 \times 10^{-3}$ C $= 9$ mC. This option is incorrect.

• heat loss is zero in this process. The initial energy stored in the system is the energy stored in the first capacitor: $U_i = \frac{1}{2} C_1 V_0^2 = \frac{1}{2} C V_0^2 = \frac{1}{2} (20 \times 10^{-6}) (1350)^2 = 10 \times 10^{-6} \times 1822500 = 18.225$ J. The final energy stored in the system is the sum of the energies stored in both capacitors: $U_f = \frac{1}{2} C_1 V_f^2 + \frac{1}{2} C_2 V_f^2 = \frac{1}{2} C V_f^2 + \frac{1}{2} kC V_f^2 = \frac{1}{2} (C+kC) V_f^2$. $C+kC = 20 \times 10^{-6} + 40 \times 10^{-6} = 60 \times 10^{-6}$ F. $U_f = \frac{1}{2} (60 \times 10^{-6}) (450)^2 = 30 \times 10^{-6} \times 202500 = 6075000 \times 10^{-6} = 6.075$ J. The heat loss is $H = U_i - U_f = 18.225 - 6.075 = 12.15$ J. Since $U_i \neq U_f$, there is heat loss. This option is incorrect.

The correct options are the first two.