CBSEGrade 12PhysicsElectrostatic Potential and Capacitance

Parallel Plate Capacitor Dilemma

A 2 μF parallel plate capacitor is charged to a potential difference of 300 V. If the distance between the plates is suddenly doubled, how does the electrostatic potential energy stored in the capacitor change, and why?

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📌 CONCEPT: The electrostatic potential energy stored in a capacitor is directly proportional to the square of the potential difference and inversely proportional to the distance between the plates.

📐 RULE / FORMULA: The formula for electrostatic potential energy stored in a capacitor is U = (1/2)CV^2, where C is the capacitance and V is the potential difference.

💡 WORKED EXAMPLE: Given a 2 μF capacitor charged to 300 V, the initial electrostatic potential energy is U1 = (1/2) * 2 * 10^-6 * (300)^2 J. If the distance between the plates is doubled, the capacitance decreases by half, and the new electrostatic potential energy is U2 = (1/2) * (1/2) * 2 * 10^-6 * (300)^2 J. Comparing U1 and U2, we see that U2 is one-fourth of U1, so the electrostatic potential energy decreases by a factor of 4.

⚠️ COMMON MISTAKE: Students often forget that capacitance is inversely proportional to distance, leading to incorrect calculations when the distance between the plates changes.

14 Jun 26

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Electrostatic Potential and Capacitance

Physics · Grade 12

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