A 6.0\text{ }\mu\text{F} capacitor is charged to a potential difference of 100\text{ V} us… — JAMB UTME

A $6.0\text{ }\mu\text{F}$ capacitor is charged to a potential difference of $100\text{ V}$ using a direct-current source and then disconnected. It is subsequently connected in parallel across an uncharged $4.0\text{ }\mu\text{F}$ capacitor. What is the total electrostatic potential energy lost in the system during the redistribution of charge, in millijoules (mJ)?

Answer: 12 mJ

Answer

The total electrostatic potential energy lost in the system during the redistribution of charge is 12 mJ.

The initial energy stored in the charged capacitor is

U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2}(6.0 \times 10^{-6}\text{ F})(100\text{ V})^2 = 30\text{ mJ}

. When connected in parallel to the uncharged capacitor, the total charge

Q = 600\text{ }\mu\text{C}

is conserved across an equivalent capacitance of

C_{eq} = 6.0\text{ }\mu\text{F} + 4.0\text{ }\mu\text{F} = 10.0\text{ }\mu\text{F}

. The common potential becomes

V_f = \frac{Q}{C_{eq}} = 60\text{ V}

, leading to a final stored energy

U_f = \frac{1}{2} C_{eq} V_f^2 = 18\text{ mJ}

. The energy lost is

\Delta U = U_i - U_f = 30\text{ mJ} - 18\text{ mJ} = 12\text{ mJ}

Step-by-Step Solution

Calculate the initial stored charge

Q

and initial energy

U_i

in the charged

6.0\text{ }\mu\text{F}

capacitor.

Q = 6.0 \times 10^{-4}\text{ C} = 600\text{ }\mu\text{C}

and

U_i = 3.0 \times 10^{-2}\text{ J} = 30\text{ mJ}

Before connection, all charge and energy reside solely on the first capacitor.

Find the equivalent capacitance

C_{eq}

when the two capacitors are connected in parallel.

C_{eq} = 6.0\text{ }\mu\text{F} + 4.0\text{ }\mu\text{F} = 10.0\text{ }\mu\text{F}

Capacitances add directly when connected in parallel.

Determine the common final potential difference

V_f

across the combination.

V_f = \frac{Q}{C_{eq}} = \frac{600\text{ }\mu\text{C}}{10.0\text{ }\mu\text{F}} = 60\text{ V}

Total electric charge is conserved during redistribution between connected capacitors.

Calculate the final total energy

U_f

stored in the combined system.

U_f = \frac{1}{2} C_{eq} V_f^2 = \frac{1}{2} (10.0 \times 10^{-6}\text{ F})(60\text{ V})^2 = 18\text{ mJ}

Both capacitors now store energy under the new common potential difference.

Compute the total energy lost

\Delta U = U_i - U_f

\Delta U = 30\text{ mJ} - 18\text{ mJ} = 12\text{ mJ}

The difference in energy is dissipated as heat in connecting wires and spark/radiation.

Key Concept

Charge Conservation and Energy Dissipation during Charge Sharing in Capacitors

Estimated Time:2m 0s

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