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A parallel plate capacitor is charged by a battery which is then disconnected. A dielectric slab is then inserted to fill the space between the plates. Match the changes that could occur with Column II and choose the correct option.
Column I Column II
A. Charge on the capacitor plates (p) Decrease by a factor of K
B. Intensity of electric field (g) Increase by a factor of K
C. Energy stored (r) Remains same
D. Capacitance (s) None

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$A \to p, B \to r, C \to q, D \to s$

$A \to p, B \to s, C \to q, D \to r$

$A \to r, B \to p, C \to p, D \to q$

$A \to r, B \to q, C \to s, D \to p$

answer is A.

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Detailed Solution

A dielectric slab is when inserted to fill the space between the plates and battery is removed, then

Quantity	Battery is removed
Capacity	C' = KC
Charge	Q' = Q
Potential	V' = V/K
Electric field	E' = E/K
Energy	U' = U/K

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