Question

A \(4~\mu F\) capacitor is charged by a \(200~V\) supply. It is then disconnected from the supply and is connected to another uncharged \(2~\mu F\) capacitors. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Answer 1

Step 1: Find initial stored energy in the capacitor}
Capacitance of the capacitor, \(C_1 = 4~\mu F\)
Supply voltage, \(V_1 = 200~V\)
Capacitance of the uncharged capacitor, \(C_2 = 2\mu F\)
Electrostatic energy stored in \(C_1\) is given as

\(E_1 = (1/2)C_1 V_1^2\)
\(= (1/2) \times 4 \times 10^{-6} \times (200)^2\)
\(= 8 \times 10^{-2}~J\)

Step 2: Apply conservation of charge
When \(C_1\) is disconnected from the power supply and connected to \(C_2\), the voltage acquired by it is \(V_2\).
According to the law of conservation of charge
\(V_2(C_1+C_2) = C_1 V_1\)
Substituting the values, we get,
\(V_2(4+2) \times 10^{-6} = 4 \times 10^{-6} \times 200\)
\(V_2 = \left ( \dfrac{400}{3} \right )V\)

Step 3: Find final stored energy in the capacitor
Electrostatic energy of the combination is
\(E_2 = (1/2)(C_1+C_2)V_1^2\)
\(= (1/2) \times (2+4) \times 10^{-6} \times (400/3)^2\)
\(=5.33 \times 10^{-2}~J\)

Step 4: Find the energy lost
Hence, amount of electrostatic energy lost by capacitor is given by,
\(C_1 = E_1 -E_2\)
\(= 0.08 - 0.0533 = 0.0267\)
\(= 2.67 \times 10^{-2}~J\)