3. A parallel plate air capaciter has a capacitance of
3 x 10-9 farad. A slab of dielectric constant 3 f thickness
3cm completely fills the space between the plates
The potential clifference between the plates is main-
tained constant at 400 volt. What is the change
in the
energy of capacitor if the slab is removeda
Ans!
Answers
Given info : A parallel plate air capaciter has a capacitance of 3 x 10^-9 Farad. A slab of dielectric constant 3, thickness 3cm completely fills the space between the plates. The potential clifference between the plates is maintained constant at 400 volt.
To find : the change in the energy of capacitor if the slab is removed, is ...
solution : charge stored in capacitor, q = CV
= 3 × 10^-9 × 400
= 12 × 10^-7 C
= 1.2 × 10^-6 C
initial capacitance, C' = KC
where K is dielectric constant, C is capacitance of capacitor.
so, C' = 3 × 3 × 10^-9 F = 9 × 10^-9 F
after removing slab between the plates, capacitance becomes C.
now, change in energy of capacitor = U - U'
= q²/2C - q²/2C'
= q²/2 [1/C - 1/C']
= (1.2 × 10^-6)²/2 [1/3 × 10^-9 - 1/9 × 10^-9 ]
= 0.72 × 10^-12 × 10^9 [1/3 - 1/9]
= 0.72 × 10^-3 × 2/9
= 0.16 × 10^-3 J
Therefore the change in energy of capacitor if slab is removed, is 0.16 × 10^-3 J