Question

A 20.0 μf capacitor is charged to a potential difference of 850 v. the terminals of the charged capacitor are then connected to those of an uncharged 12.0 μf capacitor. (a) compute the original charge of the system. (b) compute the final potential difference across capacitor. (c) compute the final energy of the system. (d) compute the decrease in energy when the capacitors are connected.

Answer 1

• (a) $Q = 1.70\times 10^{-2}\;\text{C}$;
• (b) $V_\text{final} = 5.31\times 10^{2}\;\text{V}$;
• (c) $E_\text{final} = 4.52\;\text{J}$;
• (d) $\Delta E = 2.82\;\text{J}$.

All four values are in 3 sig. fig.

Explanation

(a)

$Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}$.

(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

$C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial}$;

$(C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial}$;

$\displaystyle V_\text{final} = \frac{C_1}{C_1+C_2}\cdot V_\text{initial}\\\phantom{V_\text{final}} = \frac{20.0\;\mu\text{F}}{20.0\;\mu\text{F} + 12.0\;\mu\text{F}} \times 850\;\text{V}\\\phantom{V_\text{final}} =531\;\text{V}$.

(c)

$\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}$.

$\displaystyle E_\text{final} = \frac{1}{2} (C_1 + C_2) \cdot {V_\text{final}}^{2} \\\phantom{E_\text{final}} = \frac{1}{2} \times (20.0\times 10^{-6} + 12.0\times 10^{-6}) \times 531.25\\\phantom{E_\text{final}} = 4.52\;\text{J}$.

(d)

Initial energy of the system, which is the same as the initial energy in the $20.0\;\mu\text{F}$ capacitor:

$\displaystyle E_\text{initial} = \frac{1}{2} \times 20.0\times 10^{-6} \times 850^{2} = 7.225\;\text{J}$.

Change in energy:

$\Delta E = 7.225\;\text{J} - 4.516\;\text{J} = 2.70\;\text{J}$.