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Asked by Preeti Last Modified
Answer 
Charlie
Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C’) of the combination is given by,
New electrostatic energy can be calculated as
Therefore, the electrostatic energy lost in the process is
.
we know that 1/2 CV2 is the energy stored in a capactor with capacitance C when connected across the voltage source V.
the property of capacitor is it acquires charge when it is connected to voltage source.
after this was this connected it acts as a energy source untill the charge in it is dissipated or until an equal energy source is countered to it.
so in case 1: when it is connected to the 200V source
energy acquired by the 600pF capacitor is 1.2 * 10-5
that means the energy 1.2 *10-5 is shared between two capacitors in equal amount
that means energy lost is half of the intial total energy and that is equal to energy gained by the second capacitor = 0.6 *10-5 joules
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Case 1 - apply Q = CV
case 2- disconnected from the supply
so charge remains the same
and now capacitor is added so apply net capacitance formula for series connection of capacitor Q= C equivalent * V.'
C equivalent = net series capacitance
V' = new voltage
Dillibabu D
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Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C') of the combination is given by,
New electrostatic energy can be calculated as
Loss in electrostatic enegy = E - E'
= 1.2 x 10-5 - 0.6 x 10-5
= 0.6 x 10-5
= 6 x 10-6 J
Therefore, the electrostatic energy lost in the process is 6 x 10-6 J.
read less Jaisna Vp
Tutor
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
Capacitance of the capacitor, C = 600 pF Potential difference, V = 200 V Electrostatic energy stored in the capacitor is given by, If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C') of the combination is given by, New electrostatic energy can be calculated as Loss in electrostatic enegy = E - E' = 1.2 x 10-5 - 0.6 x 10-5 = 0.6 x 10-5 = 6 x 10-6 J Therefore, the electrostatic energy lost in the process is 6 x 10-6 J.
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Related Questions
(a) Identify an equipotential surface of the system.
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(a) Inside the sphere
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