Physics

Physics

Problem

An isolated parallel-plate capacitor with air gap dd has capacitance C0C_0, charge QQ and potential difference VV. A conducting slab of thickness d/2d/2 is inserted centrally between the plates so that it remains parallel to them and does not touch either plate. After insertion, the new potential difference between the plates is a) 2V2V b) VV c) V2\dfrac{V}{2} d) V4\dfrac{V}{4}
Intuitive explanation

When a conductor slides into a capacitor, the key insight is that the electric field inside the conductor must be zero—charges redistribute until they cancel any external field. This effectively reduces the gap where the field actually exists. Since the capacitor is isolated, the charge QQ stays constant, so you need to find how the new effective gap changes the voltage using V=EdV = Ed.

Intuitive explanation

When a conductor slides into a capacitor, the key insight is that the electric field inside the conductor must be zero—charges redistribute until they cancel any external field. This effectively reduces the gap where the field actually exists. Since the capacitor is isolated, the charge QQ stays constant, so you need to find how the new effective gap changes the voltage using V=EdV = Ed.

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