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Difficulty: MediumCapacitors and Capacitance

A capacitor of capacitance 5 μF5\text{ }\mu\text{F} is charged to a potential difference of 200 V200\text{ V} and then disconnected from the power supply. If it is subsequently connected in parallel across an uncharged capacitor of capacitance 15 μF15\text{ }\mu\text{F}, what is the final common potential difference across the combination in volts?

Answer: 50 V

Answer

The final common potential difference across the combination is 50 V.
By charge conservation, the total charge Q=C1V1=5 μF×200 V=1000 μCQ = C_1 V_1 = 5\text{ }\mu\text{F} \times 200\text{ V} = 1000\text{ }\mu\text{C} is shared across the two parallel capacitors. The total equivalent capacitance is Ctotal=C1+C2=20 μFC_{\text{total}} = C_1 + C_2 = 20\text{ }\mu\text{F}. Therefore, the final potential difference is V=QCtotal=1000 μC20 μF=50 VV = \frac{Q}{C_{\text{total}}} = \frac{1000\text{ }\mu\text{C}}{20\text{ }\mu\text{F}} = 50\text{ V}.

Step-by-Step Solution

1
Calculate the initial electric charge (QQ) stored on the charged capacitor
Q=5 μF×200 V=1000 μCQ = 5\text{ }\mu\text{F} \times 200\text{ V} = 1000\text{ }\mu\text{C}
Before connection, all charge is stored solely on the 5 μF5\text{ }\mu\text{F} capacitor.
2
Calculate the total equivalent capacitance (CtotalC_{\text{total}}) of the parallel network
Ctotal=5 μF+15 μF=20 μFC_{\text{total}} = 5\text{ }\mu\text{F} + 15\text{ }\mu\text{F} = 20\text{ }\mu\text{F}
Capacitors in parallel add directly (Ctotal=C1+C2C_{\text{total}} = C_1 + C_2).
3
Apply the law of conservation of charge to find the final common voltage (VV)
V=QCtotal=1000 μC20 μF=50 VV = \frac{Q}{C_{\text{total}}} = \frac{1000\text{ }\mu\text{C}}{20\text{ }\mu\text{F}} = 50\text{ V}
The total charge remains conserved and redistributes across the total combined capacitance.

Key Concept

Charge Redistribution and Conservation in Parallel Capacitors
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Topics

Capacitors and Capacitance

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