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Question: Plate separation of a \(15\mu F\) capacitor is 2 mm. A dielectric slab (K = 2) of thickness 1 mm is ...

Plate separation of a 15μF15\mu F capacitor is 2 mm. A dielectric slab (K = 2) of thickness 1 mm is inserted between the plates. Then new capacitance is given by

A

E0=E3E_{0} = \frac{E}{3}

B

C′=5×10=50μCC' = 5 \times 10 = 50\mu C

C

Qf=50×12=600μCQ_{f} = 50 \times 12 = 600\mu C

D

=Qf−Qi=480μC.= Q_{f} - Q_{i} = 480\mu C.

Answer

C′=5×10=50μCC' = 5 \times 10 = 50\mu C

Explanation

Solution

Given C=ε0Ad=15μFC = \frac{\varepsilon_{0}A}{d} = 15\mu F ……..(i)

Then by using C′=ε0Ad−t+tK=ε0A2×10−3−10−3+10−32=23×ε0A×103C' = \frac{\varepsilon_{0}A}{d - t + \frac{t}{K}} = \frac{\varepsilon_{0}A}{2 \times 10^{- 3} - 10^{- 3} + \frac{10^{- 3}}{2}} = \frac{2}{3} \times \varepsilon_{0}A \times 10^{3}; From equation (i) C′=20μF.C' = 20\mu F.