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Asked by Shantanu Last Modified
Answer 
Apoorva Purohit
Capacitor of the capacitance, C = 12 pF = 12 × 10−12 F
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
Therefore, the electrostatic energy stored in the capacitor is 
It can be understood by this simple formula:
energy stored in a capacitor: -
Here C = 12 pF and V = 50V
so putting into this
energy stored =
i.e. = = 15nf
Shivank Sharma
Engineer and a civil services exam aspirant.
Here, C₁ = 600pF , V₁ = 20pV , C₂ = 600pF and V₂ = 0
On connecting charged capacitor to uncharged capacitor , the common potential V across the capacitors is given by
V = (C₁V₁ + C₂V₂)/(C₁ + C₂)
= (600 × 10⁻¹² × 200 + 600 × 10⁻¹² × 0)/(600 + 600) × 10⁻¹²
= 100V
now, energy stored in capacitors before connection is
Ui = 1/2 C₁V₁² + 1/2 C₂V₂²
= 1/2 × 600 × 10⁻¹² × (200)² + 1/2 × 600 × 10⁻¹² × 0
= 12 × 10⁻⁶ J = 12μJ
And energy stored in capacitors after connection is
Uf = 1/2 (C₁ + C₂)V²
= 1/2 × (600 + 600) × 10⁻¹² × (100)²
= 1/2 × 1200 × 10⁻⁸ J
= 6 × 10⁻⁶ J
= 6μJ
Hence , energy lost in the process is ∆U = Uf - Ui
= 6μJ - 12μJ = -6μJ [ negative sign indicates energy is lost.
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Related Questions
(a) While the voltage supply remained connected.
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