Question

I have a capacitor who’s potential difference between the plates is maintained at 5V using a battery. If the capacitance is 5 F and a dielectric of relative permittivity 3 is placed between the plates such that it fully fills the space, What is the change in energy?

• A. Increases by $6.24\times {10}^{-4}J$
• B. Decreases by $6.24\times {10}^{-4}J$
• C. Increases by $9.36\times {10}^{-4}J$
• D. Decreases by $9.36\times {10}^{-4}J$

Answer 1

The correct option is A Increases by $6.24\times {10}^{-4}J$

Here the initial energy of the capacitor is
$E=\frac{1}{2}C{V}^2$
$E=\frac{1}{2}\times 5\times {10}^{-}6\times 5^2$
$E=3.12\times {10}^{-4}J$
Now when calculating the final energy, we don’t have to change V since it is maintained at 5V by means of a battery. Only Capacitance becomes C’ = kC
$E^{\prime }=\frac{1}{2}{C}^{\prime }{V}^2{E}^{\prime }=\frac{1}{2}3C{V}^2{E}^{\prime }=3E{E}^{\prime }=9.36\times {10}^{-4}J$
Energy change is, $E^{\prime }-E=6.24\times {10}^{-4}J$