Answered

An air-filled parallel-plate capacitor has plate area A and plate separation d. The capacitor is connected to a battery that creates a constant voltage V. The energy stored in the capacitor is U1. While the capacitor stays connected to the battery, a dielectric plate with thickness d and dielectric constant K, is slowly moved into the capacitor until it fills only the left half of the space between the plates. What is the new energy U2 now stored in the capacitor?

Answer :

Manetho

Answer:

[tex]U_2=\frac{1}{2}CV^2= \frac{K\epsilon AV^2}{d}[/tex]

Explanation:

We Know that electrical work done is given by

[tex]dW = V\dq[/tex]

and

now, Q=CV

C= capacitance

V= voltage from the battery

⇒V= Q/C

so, we can write

[tex]dw = \frac{q}{c}dq[/tex]

integrating you get ,

[tex]W= \frac{Q^2}{2C} =\frac{CV^2}{2} =\frac{\epsilon AV^2}{d}[/tex]

where A= area d= distance

Q does not change as battery is disconnected ,

[tex]U_1= \frac{1}{2}Q^2/C =\frac{3}{2}\frac{\epsilon V^2A}{d}[/tex]

Now put

C= K*epsilon *A/d

[tex]U_2=\frac{1}{2}CV^2= \frac{K\epsilon AV^2}{d}[/tex]

K= dielectric constant

Other Questions