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Class 12 Tuition > Learn Exercise 2 > A 600 pF capacitor is charged by a 200 V supply. It...
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A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

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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...
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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 capacitanceC= 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. read less
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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...
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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 *10-5is 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 *10-5joules read less
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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
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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 capacitanceV' = new voltage read less
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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...
read more
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 capacitanceC= 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 10-6J Therefore, the electrostatic energy lost in the process is 6 x 10-6J. read less
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Here, C₁ = 600pF , V₁ = 20pV , C₂ = 600pF and V₂ = 0On 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=...
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Here, C₁ = 600pF , V₁ = 20pV , C₂ = 600pF and V₂ = 0On 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 read less
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B.tech student,currently preparing for civil services.

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...
read more
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 capacitanceC= 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 10-6J Therefore, the electrostatic energy lost in the process is 6 x 10-6J. read less
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