Two identical air-filled parallel-plate capacitors, C_1 and C_2, each of capacitance C, ar… — JAMB UTME

Two identical air-filled parallel-plate capacitors, $C_1$ and $C_2$ , each of capacitance $C$ , are connected in series across a direct-current voltage source of potential difference $V$ . While the circuit remains connected to the voltage source, a dielectric slab of relative permittivity $\varepsilon_r = 3$ is fully inserted into $C_1$ , completely filling the space between its plates. What is the ratio of the electrostatic energy stored in $C_1$ after inserting the dielectric to the energy stored in $C_1$ before the insertion?

$3 : 4$ Answer
B
$1 : 3$
C
$3 : 1$
D
$27 : 4$

Answer

The ratio of the energy stored in the first capacitor after dielectric insertion to before insertion is

3 : 4

(or

0.75

Initially, the two identical capacitors divide the total source voltage

V

equally, giving

V_1 = V/2

and initial energy

U_{1,i} = \frac{1}{8} C V^2

. When the dielectric of relative permittivity

3

is inserted, the capacitance of the first capacitor becomes

3C

. In a series circuit connected to a constant voltage source, the total charge becomes

Q = C_{eq}V = \frac{3}{4}CV

, which reduces the voltage across the modified capacitor to

V_1' = \frac{Q}{3C} = \frac{V}{4}

. The new stored energy is

U_{1,f} = \frac{1}{2}(3C)(\frac{V}{4})^2 = \frac{3}{32} C V^2

. Dividing

U_{1,f}

U_{1,i}

yields

\frac{3/32}{1/8} = \frac{3}{4}

Step-by-Step Solution

Calculate initial capacitance and voltage across

C_1

Initial capacitance

C_{1,i} = C

. Since

C_1

and

C_2

are identical and in series, initial potential difference across

C_1

V_{1,i} = \frac{V}{2}

Equal capacitors in series divide total voltage equally.

Calculate initial electrostatic energy stored in

C_1

U_{1,i} = \frac{1}{2} C_{1,i} V_{1,i}^2 = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{8} C V^2

Formula for energy stored in a capacitor is

U = \frac{1}{2} C V^2

Determine final capacitance of

C_1

and new voltage division

New capacitance

C_{1,f} = \varepsilon_r C = 3C

. Total equivalent capacitance

C_{eq} = \frac{3C \cdot C}{3C + C} = \frac{3}{4} C

. Total charge supplied

Q = C_{eq} V = \frac{3}{4} C V

. Final voltage across

C_1

V_{1,f} = \frac{Q}{C_{1,f}} = \frac{\frac{3}{4} C V}{3C} = \frac{V}{4}

Dielectric increases capacitance by factor

\varepsilon_r

, altering equivalent capacitance and potential distribution in series.

Calculate final electrostatic energy in

C_1

and compute the ratio

U_{1,f} = \frac{1}{2} (3C) \left(\frac{V}{4}\right)^2 = \frac{3}{32} C V^2

. Ratio

\frac{U_{1,f}}{U_{1,i}} = \frac{\frac{3}{32} C V^2}{\frac{1}{8} C V^2} = \frac{3}{4}

Divide final stored energy by initial stored energy.

Key Concept

Series combination of capacitors with dielectric insertion under constant battery voltage

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