Question

A capacitor is charged to $10\mu C$ and is at potential 5V. This capacitor is now isolated from any circuit. Now a dielectric of dielectric constant k=3 is placed between the plates such that there is no empty space left between the plates. What is the change in energy of the capacitor?

• A. Increases by $7.3\times {10}^{-5}J$
• B. Decreases by $3.2\times {10}^{-5}J$
• C. Increases by $4.2\times {10}^{-5}J$
• D. Decreases by $8.3\times {10}^{-5}J$

Answer 1

The correct option is D

Decreases by $8.3\times {10}^{-5}J$

Initial energy of the capacitor is
$E=\frac{1}{2}C{V}^2$
$E=\frac{1}{2}\times 10\times {10}^{-}6\times 5^2$
$E=1.25\times {10}^{-4}J$
Once the dielectric is introduced, energy becomes
$E^{\prime }=\frac{1}{2}{C}^{\prime }{V}^{\prime 2}$
$C^{\prime }=kC$
$V^{\prime }=\frac{V}{k}$ (since V=-Ed, and E becomes $\frac{E}{k}$)
Now,
$E^{\prime }=\frac{1}{2}Ck(\frac{V}{k}{)}^2$
$E^{\prime }=\frac{E}{k}$
$E^{\prime }=4.2\times {10}^{-5}J$
Change in energy,
$E^{\prime }-E=-8.3\times {10}^{-5}J$
Energy decreases by $8.3\times {10}^{-5}J$