Question

A 10 pF parallel plate capacitor is charged with a 4.0 V battery. While the capacitor is still connected to the battery, a dielectric slab (κ = 5.0) is inserted between the plates to completely fill the gap. How much electric potential energy is stored in the capacitor after inserting the dielectric?

• A. $5.3\times {10}^{-}10J$
• B. $4.0\times {10}^{-}10J$
• C. $4.6\times {10}^{-10}J$
• D. $5.6\times {10}^{-10}J$

Answer 1

The correct option is C $4.6\times {10}^{-10}J$

$C=10pF=10\times {10}^{-12}F$

Voltage= $Q=4.0V$

Dielectric slab $(K=5.0)$

Electric potential energy= $K.q\frac{1}{2}eV{2}^2$

$=\frac{1}{2}\times 4\times 10\times {10}^{-12}$

$=20\times {10}^{-12}J$

Now, after inserting dielectric, energy available in the capacitor

$=\frac{1}{2}\times KC\times {\left(\frac{V}{K}\right)}^2$

$=\frac{1}{2}\times 5\times 10\times {10}^{-12}\times \frac{4\times 4}{5\times 5}$

$=16\times {10}^{-12}$ .

Total electric potential= $(20+16)\times {10}^{-12}$

$=46\times {10}^{-12}$

$=4.6\times {10}^{-10}J$ .