A 200 \mu F parallel plate capacitor having plate separation of 5 mm is charged by a 100 V dc source. It remains connected to the source. Using an insulated handle, the distance between the plates is doubled and a dielectric slab of thickness 5 mm and dielectric constant 10 is introduced between the plates. Explain with reason, how the (i) capacitance, (ii) electric field between the plates, (iii) energy density of the capacitor will change?

 

Answers (1)

The effective separation between the plates with air in between is 

2d-t+\frac{t}{k}

that is 

10-5+\frac{5}{10}=5.5 mm

(i)   new capacitance

C'= \frac{\epsilon_0A}{5.5mm}=\frac{10}{11}C

\approx 182\mu F

(ii) change in the electric field or effective new electric field

=\frac{100v}{5.5\times 10^{-3}m}

\approx 18182\; v/m

(iii) Energy density

Energy=\frac{1}{2}CV^2

As C has decreased energy also decreases, so energy density also decreases.

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