Question 21: Electrostatics & Capacitance

Question

A dielectric slab of dielectric constant $k=3$ is filling $\frac{3}{4}$ the space of the capacitor.
When capacitor is charged, then % of total energy stored in dielectric is
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Accepted Answer

Energy density in air gap $=\frac{1}{2} \varepsilon_0 E^2=u_E$ $\$
Energy density of dielectric $=\frac{1}{2} \varepsilon_0 K \frac{E^2}{K^2}=\frac{u_E}{K}$ $\$
Now, energy stored $=$ energy density $	imes$ volume $\$
Total energy stored
$=\frac{3}{4} \cdot x \cdot \frac{u_E}{K}+\frac{1}{4} \cdot x \cdot u_E[x=$ volume of capacitor $]$
$=x u_E\left(\frac{3}{4 K}+\frac{1}{4}ight) 	herefore K=3$; $\$
$=x u_E\left(\frac{2}{4}ight)=\frac{x u_E}{2}$ $\$
Energy stored in dielectric $=\frac{3}{4} \cdot x \cdot \frac{u_E}{K}=\frac{x u_E}{4}$ $\$
So, % of energy stored in the dielectric $\$
$ =\frac{\frac{x u_E}{4}}{\frac{x u_E}{2}} 	imes 100=50 \% $

Answer

25%

Answer

50%

Answer

75%

Answer

12.5%

Electrostatics & Capacitance

A dielectric slab of dielectric constant k=3 is filling \frac{3}{4} the space of the capacitor. When capacitor is charged, then % of total energy stored in dielectric is

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