CBSEGrade 12PhysicsElectrostatic Potential and Capacitance

Capacitance of a Parallel-Plate Capacitor?

A parallel-plate capacitor is made of two plates, each 5 cm x 10 cm in size, separated by a 2 mm thick dielectric material. The plates are 1 cm apart when empty. How will the capacitance change if the gap between the plates is reduced to 1 mm and the dielectric material is replaced with a different one of double the dielectric constant?

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📌 CONCEPT: Capacitance is the ability of a capacitor to store electric charge, measured in Farads (F).

📐 RULE / FORMULA: The capacitance (C) of a parallel-plate capacitor is given by the formula C = ε₀εᵣ(A/d), where ε₀ is the permittivity of free space, εᵣ is the relative permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates.

💡 WORKED EXAMPLE: If the original capacitance of the capacitor is 2.5 x 10⁻⁹ F and the area of the plates is 50 cm², what is the new capacitance if the distance between the plates is reduced to 1 mm and the dielectric material is replaced with a material of double the dielectric constant?

⚠️ COMMON MISTAKE: Students often forget to account for the change in dielectric constant when replacing the dielectric material, which can lead to incorrect capacitance values.

22 Sept 26