Question 23.9: A capacitor C1 = 4µF is charged by an initial potential diff......
A capacitor C_{1} = 4µF is charged by an initial potential difference ΔV_{i} = 12 V, see Fig. 23.15a. The charging battery is then removed, as shown in Fig. 23.15b, and the capacitor is connected to the uncharged capacitor C_{2} = 2 µF, as shown in Fig. 23.15c. (a) Find the final potential difference ΔV_{f} as well as Q_{1f} and Q_{2f}. (b) Find the stored energy before and after the switch is closed.
(a) The original charge is now shared by C_{1} and C_{2}, so:
Q_{1i}=Q_{1f}+Q_{2f}Using of the relation Q = CΔV in each term of this equation, we get:
C_{1}\Delta V_{\mathrm{i}}=C_{1}\Delta V_{\mathrm{f}}+C_{2}\Delta V_{\mathrm{f}}
Thus:
\Delta V_{\mathrm{f}}=\frac{C_{1}}{C_{1}+C_{2}}\Delta V_{\mathrm{i}}=\frac{(4 \ \mu{F})}{4\ \mu\mathrm{F}+2\ \mu{F}}(12\,\mathrm{V})=8\,\mathrm{V}
Q_{1\mathbf{f}}=C_{1}\Delta V_{{f}}=(4\ \mu F)(8 \ V)=32\operatorname{\mu C}
and
Q_{2\mathbf{f}}=C_{2}\Delta V_{{f}}=(2\ \mu F)(8 \ V)=16\operatorname{\mu C}
(b) The initial potential energy is:
U_{\mathrm{i}}={\textstyle\frac{1}{2}}C_{1}(\Delta V_{\mathrm{i}})^{2}={\textstyle\frac{1}{2}}(4 \ \mu F)(12\ V)^{2}=288 \ \mu JThe final potential energy is:
U_{{f}}={\textstyle\frac{1}{2}}C_{1}(\Delta V_{{f}})^{2}+{\textstyle\frac{1}{2}}C_{2}(\Delta V_{{f}})^{2}={\textstyle\frac{1}{2}}(4\ \mu {F}+2\ \mu {F})(8 \ {V})^{2}=192 \ \mu{J}Although U_{i} > U_{f}, this is not a violation of the conservation of energy principle.
The missing energy is transferred as thermal energy into the connecting wires and as radiated electromagnetic waves.