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>> Capacitors and Energy Interpretation
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’