Question

Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V, its capacitance is : (in pF)

• A. 600
• B. 200
• C. 400
• D. 100

Hint:

When a charged capacitor is connected in parallel to an uncharged capacitor, charge is redistributed until the voltage across both capacitors is the same. The total charge in the system is conserved. You can use the conservation of charge or the formula for the final voltage in parallel capacitor combinations to solve such problems.

Answer 1

The Correct Option is B

Let the capacitance of the first capacitor be \( C_1 = 100 \, \text{pF} \) and its initial voltage be \( V_i = 60 \, \text{V} \). 
The initial charge on the first capacitor is \( Q_i = C_1 V_i = (100 \, \text{pF})(60 \, \text{V}) = 6000 \, \text{pC} \). 
A second uncharged capacitor with capacitance \( C_2 \) is connected in parallel to the first capacitor. When capacitors are connected in parallel, the voltage across them becomes equal. 
The final voltage across the second capacitor is given as \( V_f = 20 \, \text{V} \). 
Since they are in parallel, the final voltage across the first capacitor is also \( V_f = 20 \, \text{V} \). 
The total charge in the system is conserved. The initial charge was only on the first capacitor, \( Q_i = 6000 \, \text{pC} \). 
After connecting the second capacitor, this charge is distributed between the two capacitors. 
The final charge on the first capacitor is \( Q_{f1} = C_1 V_f = (100 \, \text{pF})(20 \, \text{V}) = 2000 \, \text{pC} \). 
The final charge on the second capacitor is \( Q_{f2} = C_2 V_f = C_2 (20 \, \text{V}) \). 
By conservation of charge: \[ Q_i = Q_{f1} + Q_{f2} \] \[ 6000 \, \text{pC} = 2000 \, \text{pC} + C_2 (20 \, \text{V}) \] \[ 4000 \, \text{pC} = C_2 (20 \, \text{V}) \] \[ C_2 = \frac{4000 \, \text{pC}}{20 \, \text{V}} = 200 \, \text{pF} \] The capacitance of the second capacitor is 200 pF. Alternatively, using the formula for the final voltage when a charged capacitor \( C_1 \) with initial voltage \( V_i \) is connected in parallel to an uncharged capacitor \( C_2 \): \[ V_f = \frac{C_1 V_i}{C_1 + C_2} \] Given \( V_f = 20 \, \text{V} \), \( C_1 = 100 \, \text{pF} \), and \( V_i = 60 \, \text{V} \): \[ 20 = \frac{(100)(60)}{100 + C_2} \] \[ 20 (100 + C_2) = 6000 \] \[ 2000 + 20 C_2 = 6000 \] \[ 20 C_2 = 4000 \] \[ C_2 = \frac{4000}{20} = 200 \, \text{pF} \]

Answer 2

To solve this problem, we need to analyze the given capacitor setup and use the appropriate formulas for capacitors in parallel circuits.

Initial Setup:

• We have a 100 pF capacitor charged to 60 V. Let's denote this capacitor as \( C_1 \).
• The initial charge on \( C_1 \) can be calculated using the formula: \(Q = C \times V\)
• So, the charge on \( C_1 \) is: \(Q_1 = 100 \, \text{pF} \times 60 \, \text{V} = 6000 \, \text{pC} \, (\text{picoCoulombs})\)

After Connecting the Second Capacitor:

• Once the second uncharged capacitor, \( C_2 \), is connected in parallel with \( C_1 \), the total charge is conserved during redistribution.
• The final voltage across both capacitors is 20 V.
• The total charge initially is solely that on \( C_1 \), which is 6000 pC.
• After redistribution, the total charge is still 6000 pC, now across the total capacitance of the two capacitors in parallel: \(Q_{\text{total}} = (C_1 + C_2) \times 20 \, \text{V}\)
• Therefore, we have: \(6000 \, \text{pC} = (100 \, \text{pF} + C_2) \times 20 \, \text{V}\)

Solving for \( C_2 \):

• Simplify the equation: \(6000 = 20(100 + C_2)\)
• Divide both sides by 20: \(300 = 100 + C_2\)
• Solve for \( C_2 \): \(C_2 = 300 - 100 = 200 \, \text{pF}\)

Conclusion:

• The capacitance of the second capacitor is 200 pF, which corresponds to the correct answer of \(200\).

Parameters	Values
Initial Voltage on \( C_1 \)	60 V
Capacitance of \( C_1 \)	100 pF
Final Voltage	20 V
Charge Conservation	\(6000 \, \text{pC} = (100 + C_2) \times 20\)
Capacitance of \( C_2 \)	200 pF