Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:

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Accepted Answer

(4) 0.5×10⁻⁶ J. Energy lost when two capacitors share charge: Δ U=1/2(C₁C₂)/(C₁+C₂)(V₁-V₂)². (C₁C₂)/(C₁+C₂)=(200×200)/(400)=100 pF =10⁻¹⁰ F. Δ U=1/2×10⁻¹⁰×(100)²=0.5×10⁻⁶ J. Check: Uᵢ=1/2×2×10⁻¹⁰×10⁴=10⁻⁶ J; after sharing each has 50 V, U_f=2×1/2×2×10⁻¹⁰×2500=0.5×10⁻⁶ J.

Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:

Step-by-step solution

Why the other options are wrong