Question

A 2.50-nF parallel-plate capacitor is charged to an initial potential difference ΔVi = 100 V and is then isolated. The dielectric material between the plates is paper, with a dielectric constant of 3.7.

Requried:
a. How much work is required to withdraw the mica sheet?
b. What is the potential difference across the capacitor after the mica is withdrawn?

Answer 1

Step-by-step explanation:

Formula to calculate the energy stored in a capacitor when it is filled with air is,

`U_(1)=(1)/(2) C V_(1)^(2)`]

Here,
`U_(1)` is the energy stored in a capacitor when it is filled with air.

`C` is the parallel plate capacitor.

[/tex]V_{\mathrm{i}}[/tex] is the initial potential difference.

Substitute
`2.00 \mathrm{nF}` for
`C` and
`100 \mathrm{V}` for
`V_{\mathrm{i}}` to find the
`U_(1)`

`\begin{array}{c} U_(1)=(1)/(2)\left(2.00 \mathrm{nF}\left(\frac{10^(9) \mathrm{F}}{1 \mathrm{n} \mathrm{F}}\right)\right)(100 \mathrm{V})^(2) \\ =10^(-5) \mathrm{J} \end{array}`]

Formula to calculate the energy stored in a capacitor when it is filled with dielectric is,