A \(5 \mu \mathrm{F}\) capacitor is charged by a \(220 \mathrm{v}\) supply. It is then disconnected from the supply and is connected to another uncharged $2.5 \mu \mathrm{F}$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ? (A) \(0.02 \mathrm{~J}\) (B) \(0.121 \mathrm{~J}\) (C) \(0.04 \mathrm{~J}\) (D) \(0.081 \mathrm{~J}\)

Final Voltage = 146.67 V #tag_title#Step 3: Calculate the final energy stored in the connected capacitors#tag_content#Now that we have the final voltage across the connected capacitors, we can calculate the final energy stored in them using the energy stored in a capacitor formula: Final Energy = \(\frac{1}{2}\) × Total Capacitance × Final Voltage^2 Final Energy = \(\frac{1}{2}\) × 7.5 × 10^{-6} F × (146.67 V)^2 #tag_title#Step 4: Determine the energy lost#tag_content#Finally, we can find the energy lost in the process by subtracting the final energy stored in the connected capacitors from the initial energy stored in the charged capacitor: Energy Lost = Initial Energy - Final Energy Energy Lost = \(\frac{1}{2} \times 5 \times 10^{-6} \text{F} \times (220 \text{V})^2 - \frac{1}{2} \times 7.5 \times 10^{-6} \text{F} \times (146.67 \text{V})^2\) Energy Lost = 0.121 J The energy lost in the form of heat and electromagnetic radiation during the process is (B) 0.121 J.