Changing the plate spacing of a parallel plate capacitor and electrical energy

Explanation

Initial State and Changed Capacitance and Charge

Let $C_1$C1​ be the capacitance of the capacitor when the plate spacing is $d$d, and $Q_1$Q1​ be the magnitude of the stored charge. $C_1 = \varepsilon_0 \frac{S}{d}$C1​=ε0​dS​ $Q_1 = C_1 V = \varepsilon_0 \frac{SV}{d}$Q1​=C1​V=ε0​dSV​

Let $C_2$C2​ be the capacitance and $Q_2$Q2​ be the magnitude of the stored charge when the plate spacing is increased to $3d$3d. Since it remains connected to the power supply, the voltage $V$V is constant.

$C_2 = \varepsilon_0 \frac{S}{3d} = \frac{1}{3} C_1$C2​=ε0​3dS​=31​C1​ $Q_2 = C_2 V = \varepsilon_0 \frac{SV}{3d} = \frac{1}{3} Q_1$Q2​=C2​V=ε0​3dSV​=31​Q1​

Work done by the power supply

By widening the gap between the plates, the charge on the capacitor decreases from $Q_1$Q1​ to $Q_2$Q2​. This decreased charge $\Delta Q$ΔQ flows back from the positive plate to the positive terminal of the power supply. $\Delta Q = Q_1 - Q_2 = \frac{2}{3} Q_1 = \frac{2\varepsilon_0 SV}{3d}$ΔQ=Q1​−Q2​=32​Q1​=3d2ε0​SV​

Because the charge flows into the positive terminal against the voltage $V$V of the power supply, the power supply absorbs electrical energy. Therefore, the work done by the power supply $W$W can be calculated as follows:

$W = \Delta Q \times V = \frac{2\varepsilon_0 SV^2}{3d}$W=ΔQ×V=3d2ε0​SV2​

Numerical Substitution

Substitute the given constraint values ​​into the equation. $W = \frac{2 \times (8.85 \times 10^{-12}) \times 0.200 \times 300^2}{3 \times (2.00 \times 10^{-3})}$W=3×(2.00×10−3)2×(8.85×10−12)×0.200×3002​

Rearrange the values ​​in the numerator and denominator and perform the calculation.

• $V^2 = 300^2 = 9.00 \times 10^4$V2=3002=9.00×104
• Numerator: $2 \times 8.85 \times 10^{-12} \times 0.200 \times 9.00 \times 10^4 = 31.86 \times 10^{-8} = 3.186 \times 10^{-7}$2×8.85×10−12×0.200×9.00×104=31.86×10−8=3.186×10−7
• Denominator: $3 \times 2.00 \times 10^{-3} = 6.00 \times 10^{-3}$3×2.00×10−3=6.00×10−3

From these, we calculate the work $W$W.

$W = \frac{3.186 \times 10^{-7}}{6.00 \times 10^{-3}} = 0.531 \times 10^{-4} = 5.31 \times 10^{-5} \text{ J}$W=6.00×10−33.186×10−7​=0.531×10−4=5.31×10−5 J

The input format is $A \times 10^{-7} \text{ J}$A×10−7 J, so we adjust the number of decimal places.

$5.31 \times 10^{-5} = 531 \times 10^{-7} \text{ J}$5.31×10−5=531×10−7 J

Therefore, the coefficient $A$A is $531$531.

Answer: 531