Chapter 19, Problem 78GP

To determine

The value of the new capacitance.

Answer to Problem 78GP

The value of the new capacitance is $C_0\left(\frac{K_1+K{2}_{}}{2}\right)$ .

Explanation of Solution

Given information:

The capacitance of the air-filled parallel plate capacitor is $C_0$ . The dielectric constant of two identical dielectric labs are $K_1$ and $K_2$ .

Formula used:

Let the area of the parallel plate is $A$ and the separation distance is $d$ . Write the expression to calculate the capacitance of the parallel plate.

  $C_0=\in \frac{A}{d}$    … (1)

Calculation:

Let two identical dielectric slabs are inserted. Let the capacitance of the identical slabs are $C{1}_{}$ and $C{2}_{}$ .

Write the expression to calculate the capacitance of the dielectric slabs.

  $C_1=K_1\in \frac{\frac{A}{2}}{d}$

and,

  $C_2=K_2\in \frac{\frac{A}{2}}{d}$

Since the slabs are connected in parallel. So, the new capacitance is the sum of the capacitance of the dielectric slabs.

  $C=C_1+C_2$

Where,

• $C$is the new capacitance.

Substitute the values in the above expression.

  $\begin{array}{c}C=K_1\in \frac{\frac{A}{2}}{d}+K_2\in \frac{\frac{A}{2}}{d}\\ =\in \frac{A}{d}\left(\frac{K_1+K{2}_{}}{2}\right)\\ =C_0\left(\frac{K_1+K{2}_{}}{2}\right)\end{array}$

Therefore, the value of the new capacitance is $C_0\left(\frac{K_1+K{2}_{}}{2}\right)$ .