A 200 microfarad parallel plate capacitor having plate separation of 5mm is charged by a100v dc source.It remains connected to the source. Using an insulted handle, the distance between the plates is doubles and a dielectric sab of thickness 5mm and dielectric constant 10is introduced between the pl
Answer (1)
Student Expert
14th Jul, 2021
Hello, i am here to help you according to your query.
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.
Hope,this information will helpful to you.
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