A 100 \mu F parallel plate capacitor having plate separation of 4 mm is charged by 200 V dc. The source is now disconnected. When the distance between the plates is doubled and a dielectric slab of thickness 4 mm and dielectric constant 5 is introduced between the plates, how will (i) its capacitance, (ii) the electric field between the plates, and (iii) energy density of the capacitor get affected? Justify your answer in each case.

 

Answers (1)
S safeer

The effective separation between the plates with air in between is 

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

that is 

8-4+\frac{4}{5}=4.8 mm

(i)   new capacitance

C'= \frac{\epsilon_0A}{4.8 mm}=\frac{4}{4.8}C

= 83.33\mu F

(ii) The charge remains the same. There fore new voltage 

V'=\frac{Q}{C'}=\frac{100\times 10 ^{-6}\times200}{83.33\times 10^{-6}}=240V

Electric field

E'=\frac{V'}{24}=30N/C

(iii) Energy density

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

As the electric field has decreased the energy density also decreases since the energy density 

U=\frac{1}{2}\epsilon_0E^2 

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