TRANSFERRING CHARGE AND ENERGY BETWEEN CAPACITORS We connect a capacitor C1 = 8.0 μF to a power supply, charge it to a potential difference V0 = 120 V, and disconnect the power supply (Fig. 24.12). Switch S is open. (a) What is the charge Q0 on C1? (b) What is the energy stored in C1?

Question

TRANSFERRING CHARGE AND ENERGY BETWEEN CAPACITORS

We connect a capacitor [latex]C_1[/latex] = 8.0 μF to a power supply, charge it to a potential difference [latex]V_0[/latex] = 120 V, and disconnect the power supply (Fig. 24.12). Switch S is open. (a) What is the charge [latex]Q_0[/latex] on [latex]C_1[/latex]? (b) What is the energy stored in [latex]C_1[/latex]? (c) Capacitor [latex]C_2[/latex] = 4.0 μF is initially uncharged. We close switch S. After charge no longer flows, what is the potential difference across each capacitor, and what is the charge on each capacitor? (d) What is the final energy of the system?

Accepted Answer

IDENTIFY and SET UP:

In parts (a) and (b) we find the charge [latex]Q_0[/latex] and stored energy [latex]U_{initial}[/latex] for the single charged capacitor [latex]C_1[/latex] from Eqs. (24.1) and (24.9), respectively. After we close switch S, one wire connects the upper plates of the two capacitors and another wire connects the lower plates; the capacitors are now connected in parallel. In part (c) we use the character of the parallel connection to determine how [latex]Q_0[/latex] is shared between the two capacitors. In part (d) we again use Eq. (24.9) to find the energy stored in capacitors [latex]C_1[/latex] and [latex]C_2[/latex]; the energy of the system is the sum of these values.

[latex]C=\frac{Q}{V_{a b}}[/latex] (24.1)

[latex]U=\frac{Q^{2}}{2 C}=\frac{1}{2} C V_{ }^{2}=\frac{1}{2} Q V[/latex] (24.9)

EXECUTE:

(a) The initial charge [latex]Q_0[/latex] on [latex]C_1[/latex] is

[latex]Q_{0}=C_{1} V_{0}=(8.0 \mu \mathrm{F})(120 \mathrm{~V})=960 \mu \mathrm{C}[/latex]

(b) The energy initially stored in [latex]C_1[/latex] is

[latex]U_{ ext {initial }}=\frac{1}{2} Q_{0} V_{0}=\frac{1}{2}\left(960 imes 10^{-6} \mathrm{C} ight)(120 \mathrm{~V})=0.058 \mathrm{~J}[/latex]

(c) When we close the switch, the positive charge [latex]Q_0[/latex] is distributed over the upper plates of both capacitors and the negative charge -[latex]Q_0[/latex] is distributed over the lower plates. Let [latex]Q_1[/latex] and [latex]Q_2[/latex] be the magnitudes of the final charges on the capacitors. Conservation of charge requires that [latex]Q_1[/latex] + [latex]Q_2[/latex] = [latex]Q_0[/latex]. The potential difference V between the plates is the same for both capacitors because they are connected in parallel, so the charges are [latex]Q_1[/latex] = [latex]C_1[/latex]V and [latex]Q_2[/latex] = [latex]Q_2[/latex]V.
We now have three independent equations relating the three unknowns [latex]Q_1[/latex], [latex]Q_2[/latex], and V. Solving these, we find

[latex]\begin{aligned}V &=\frac{Q_{0}}{C_{1}+C_{2}}=\frac{960 \mu \mathrm{C}}{8.0 \mu \mathrm{F}+4.0 \mu \mathrm{F}}=80 \mathrm{~V} \Q_{1} &=640 \mu \mathrm{C} \quad Q_{2}=320 \mu\mathrm{C}\end{aligned}[/latex]

(d) The final energy of the system is

[latex]\begin{aligned}U_{ ext {final }} &=\frac{1}{2} Q_{1} V+\frac{1}{2} Q_{2} V=\frac{1}{2}Q_{0} V \&=\frac{1}{2}\left(960 imes 10^{-6} \mathrm{C} ight)(80 \mathrm{~V})=0.038\mathrm{~J}\end{aligned}[/latex]

EVALUATE: The final energy is less than the initial energy; the difference was converted to energy of some other form. The conductors become a little warmer because of their resistance, and some energy is radiated as electromagnetic waves. We’ll study the behavior of capacitors in more detail in Chapters 26 and Chapter 31.

Question 24.7: TRANSFERRING CHARGE AND ENERGY BETWEEN CAPACITORS We connect...

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