Energy Stored in a Capacitor

Energy supplied by Source

Energy lost in form of heat =

Energy density of the electric field=

Force on a Dielectric in a Capacitor

Example 11. A 4μf capacitor is charged to 150 V and another 6μf capacitor is charged to 200 V. Then they are connected across each other. Find the potential difference across them. Calculate the heat produced.

Solution :    4μf charged to 150 V would have q₁ = C₁V₁ = 600μC
    6μf charged to 200 V would have q₂ = C₂V₂ = 1200μC
    After connecting them across each other, they will have a common potential difference V.
    Charges will readjust as q₁’ and q₂’