A parallel plate air capacitor of capacitance 8.0\text{ }\mu\text{F} is fully charged usin… — JAMB UTME

A parallel plate air capacitor of capacitance $8.0\text{ }\mu\text{F}$ is fully charged using a $40.0\text{ V}$ d.c. power supply and then disconnected from the source. A dielectric slab with a relative permittivity of $4.0$ is subsequently inserted to completely fill the region between the plates. Calculate the magnitude of the decrease in electrostatic energy stored in the capacitor, in microjoules ( $\mu\text{J}$ ).

Answer: 4800 μJ

Answer

The magnitude of the decrease in stored electrostatic energy is 4800 μJ.

Disconnecting the battery ensures that the charge

Q = C_0 V = 320\text{ }\mu\text{C}

on the plates stays fixed. Inserting the dielectric increases capacitance fourfold to

32.0\text{ }\mu\text{F}

. The energy decreases from

U_i = 6400\text{ }\mu\text{J}

U_f = \frac{Q^2}{2C_f} = 1600\text{ }\mu\text{J}

, giving a total decrease of

4800\text{ }\mu\text{J}

Step-by-Step Solution

Calculate the initial energy stored in the air capacitor before disconnection.

Initial energy

U_i = \frac{1}{2} C_0 V^2 = \frac{1}{2} \times 8.0 \times 10^{-6} \text{ F} \times (40.0 \text{ V})^2 = 6.4 \times 10^{-3} \text{ J} = 6400 \text{ } \mu\text{J}

The initial state has known capacitance and potential difference.

Calculate the new capacitance with the dielectric present.

Final capacitance

C_f = K C_0 = 4.0 \times 8.0 \text{ } \mu\text{F} = 32.0 \text{ } \mu\text{F}

Inserting a dielectric of constant

K

scales the capacitance by

K

Calculate the final stored energy using charge conservation.

Final energy

U_f = \frac{U_i}{K} = \frac{6400 \text{ } \mu\text{J}}{4.0} = 1600 \text{ } \mu\text{J}

Disconnection forces the charge

Q

to remain fixed, so energy scales inversely with capacitance (

U = \frac{Q^2}{2C}

Find the difference between initial and final energy.

\Delta U = 6400 \text{ } \mu\text{J} - 1600 \text{ } \mu\text{J} = 4800 \text{ } \mu\text{J}.

The loss in electrostatic energy represents the work done by the field pulling the dielectric slab into the plates.

Key Concept

Effect of dielectric insertion on stored electrostatic energy under isolated (constant charge) conditions

Estimated Time:2m 0s

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