Question

A 200 uF parallel plate capacitor having plate separation of 5mm ias chrged by a 100 V dc source. It remains connected to the source. Using an insulated handle, the distance between the plates is doubled and a dielactric slab of thickness 5mm and dielectric constant 10 is introduced between the plates. Explain with reason, how the
1 )capaciance
2) electric fieret hote the plate
3)energy density of raperitar will change?​

Answer 1

answer to hi electric fieret hote the plate

Answer 2

Given: Capacitance, C = 200micro Farad

Potential Difference, V = 100V

Dielectric Constant, k = 10

thickness, t = 5mm

plate separation  distance, d = 5mm

Find:

a) Capacitance

b) Electric Field

c) Energy Density

Step by Step Solution:

First of all we calculate the effective separation between the plates with air in between is

$2d-t +\dfrac{t}{k} \\\\2\times5- 5 +\dfrac{5}{10}\\\\10-5+0.5\\5.5 mm$

Now, we calculate the new capacitance,

a)

$C' = \dfrac{\epsilon_{0}A }{5.5}\\\\C' = \dfrac{10}{11}\times C \\\\C' = \dfrac{10}{11} \times 200\\\\C' = 181.81 \mu F$

b)

$E = \dfrac{V}{d} \\\\E = \dfrac{100}{5.5\times 10^{-3} }\\\\E = 1818.18 V/m$

c)

$Energy = \dfrac{1}{2} CV^{2}$

As Capacitance has decreased energy also decreases, so energy density also decreases.