Two identical capacitors are charged to different potentials `φ_(1) and φ_(2)` relative to the negative earthed electrodes. The capacitors are then connected in parallel (Fig. 24.21). Find the potential of the battery after the connection was made and the change in the energy of the system.

Explanation

Before the connection is made there is a charge `q_(1)=C varphi_(1)` on the first capacitor, and `q_(2)=C varphi_(2)`, on the second. After the upper plates of the capacitors are connected, the charge `q=q_(1)+q_(2)` is equipartitioned between them. The potential of the unearthed plates is
`varphi_("sys")=(q)/(C_("sys"))=(q_(1)+q_(2))/(2C)=(varphi_(1)+varphi_(2))/(2)`
The energy of the system before the connection is
`W_("sys")=(C_(svs) varphi_("sys")^(2))/(2)=(2C(varphi_(1)+varphi_(2))^(2))/(2xx4)=(C(varphi_(1)+varphi_(2))^(2))/(4)`
This is less than before the connection. The lost energy is transformed into other types of energy (heating the conductors, forming a spark, electromagnetic radiation, etc.)