Answer to Question #86260 in Electric Circuits for SHAHI

Question

Is there any changes occur when a dielectric slab of thickness d/2 and separation between the plates is d, if intially slab at one corner plate (slab stay attached with the plate )
If we slowly moving the slab to the other corner of the plate, work done in moving the slab, dielectric constant is K

Answer 1

Yes, the will be changes in capacitance and voltage (if it was charged) when you slowly insert the dielectric slab into the capacitor. You can consider the 'empty' capacitor having capacitance

$$C_1=\frac{\epsilon_0 A}{d},$$

and the capacitor with the slab inside as two capacitors: empty with d/2 and with the dielectric d/2-thick slab, and the final capacitance with the dielectric slab fully inside the capacitor will be (since these two capacitors are "connected" in series):

$$\frac{1}{C_2}=\frac{1}{\frac{\epsilon_0 A}{d/2}}+\frac{1}{\frac{K \epsilon_0 A}{d/2}}.$$

Now determine the work required to slowly move the slab from one to the other corner of the plate. Let's imagine that the slab is $x$ m inside the plates, and the length of the plates is $L$ :

We have three capacitors:

1) blue: thickness: d/2, with dielectric K, connected in series with 2):

2) above the blue one: d/2, K=1. 1) and 2) are connected in parallel with 3):

3) at their left: thickness d, K=1, connected in parallel with 1) and 2).

The capacitance for any $x$ can be expressed as

$$C=\frac{\epsilon_0 A(L-x)}{Ld}+\frac{2\epsilon_0 KAx}{Ld(K+1)}.$$

The work done by the external force equals the difference between the initial and final energy of the capacitor. Since in the very beginning the dielectric was on the one side of the capacitor, and in the end it was $L$ meters further, the initial and final energies are equal, and the work is 0.