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The separation between the plates of a parallel-plate capacitor of capacitance `2muF` is 2mm. The capacitor is initially uncharged. If a dielectric slab of surface area same as the capacitor, thickness 1mm, and dielectric constant 3 is introduced, and then the capacitor is charged to a potential difference difference 100V, the energy stored in the capacitor becomes _______mJ.

Text Solution

Verified by Experts

The correct Answer is:
15
Let the plate separation be d, the plate area be A, and the initial capacitance be `C_(0)`.
Then, `C_(0)=(epsi_(0)A)/(d)`
After the dielectric slab is introduced, let the capacitance be C.
So, `(1)/(C)=(((d)/(2)))/(epsi_(0)A)+(((d)/(2)))/(3psi_(0)A) implies C=(3C_(0))/(2)`.
Since `C_(0)=2muF, C=3muF`
So, energy stored in the capacitor, `U=(1)/(2)CV^(2)=(1)/(2)(3xx10^(-6))(100)^(2)=15mJ`.
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