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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) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
(a) While the voltage supply remained connected.
(b) After the supply was disconnected.
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