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  4. The energy and capacity of a charged parallel plate capacitor are U and C respectively. Now a dielectric slab of εr=6 is inserted in it then energy and capacity becomes (Assuming charge on plates remains constant)

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Q. The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of $\varepsilon_{r}=6$ is inserted in it then energy and capacity becomes (Assuming charge on plates remains constant)

Electrostatic Potential and Capacitance

Solution:

$C '=\frac{\epsilon_{0} \epsilon_{r} A }{d}$
$ \Rightarrow C '=6 C$
$U =\frac{q^{2}}{2 C}$
$\Rightarrow U '=\frac{q^{2}}{2 \times C'}=\frac{q^{2}}{2 \times 6 C}$
$=\frac{U}{6}$