Question

Capacitance of a capacitor is 16 uF. If a dielectric slab of dielectric constant 8 is inserted between the plates of capacitor.
Then the capacitance of the capacitor will be
A) 128uF
B) 16uF
C) None of these
D) 2uF​

Answer 1

B is correct

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Explanation

Given, 16μF V = Q/C = Q/16μF When dielectric is inserted V' = Q/C = Q/K (16μF) = V/8 Q/K (16μF) = Q/8 × 16μF K = 8 Option B is correct.