Question

A parallel plate capacitor of capacitance c is charged to a potential v by a battery. Without disconnecting the battery,the distance between the plates is tripled and a dielectric medium of k=10 is introduced between the plates of the capacitor. Explain giving reasons how would i) charge on the capacitor

Answer 1

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Answer 2

Answer:

Energy density becomes 40 times that of the initial energy density.

Explanation:

The potential of a parallel plate capacitor is charged to V. After a while, the distance between the plates is reduced, and a k=10 dielectric medium is added without cutting off the DC source.

When the DC source remains connected, potential across the plates remains the same.

1. Initial capacitance,

$\mathrm{C}=\frac{\in_0 A}{d}$

When the spacing between the plate is halved $d=d / 2$

A dielectric slab of $\mathrm{K}=10$ is inserted.

New capacitance,

$\mathrm{C}^{\prime}=k \frac{\in_0 A}{d / 2}=10$

$\frac{2 \in_0 A}{d}=20 C$

New capacitance becomes 20 times that of the initial capacitance.

2. Electric field is given by

$\mathrm{E}=\frac{V}{d}$

When spacing is halved

New electric field becomes

$\mathrm{E}^{\prime}=\frac{2 V}{d}=2 \mathrm{E}$

I heretore, the electric tield becomes twice that of the initial tield.

3. Initial energy density $=\frac{1}{2} \in_0 E^2$

$\mathrm{U}=\frac{1}{2} \in_0 E^2$

A dielectric slab of $\mathrm{K}=10$ is inserted.

New energy density

$U^{\prime}=\frac{1}{2} K \in_0(2 E)^2$

$\begin{aligned}& =\frac{1}{2}(10) \epsilon_0 \times 4 E^2 \\& =40\left(\frac{1}{2} \in_0 E^2\right) \\& =40 \mathrm{U}\end{aligned}$

Therefore, energy density becomes 40 times that of the initial energy density.

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