A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

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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
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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 read less
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Engineer and a civil services exam aspirant.

E=1/2{CV^2} =1/2× {(12×10^-2)×(50^2) } =1.5×10^-8 joule
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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 [ negative sign indicates energy is lost. read less
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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
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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 read less
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Famous UNACADEMY PHYSICS TEACHER (use code ROHIT10 for 10% discount on unacademy)

Apply the formula of energy stored in the capacitor i.e Energy(U) = 1/2 C V(square) = 0.5(12x10(power -12)).(50) = 3x 10(power -10)
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Perfect learning for Accounts and Astrology

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
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