Understanding the behavior of a parallel plate capacitor when a dielectric slab is inserted is crucial in electromagnetism. This problem involves calculating the new capacitance of a parallel plate capacitor after a dielectric material is introduced between its plates. We will determine the new capacitance, often denoted as C', in terms of the original capacitance \(C_0\).

Original Capacitance of the Parallel Plate Capacitor

Initially, we consider a parallel plate capacitor with plate area 'A' and separation 'd'. The space between the plates is filled with vacuum or air. The original capacitance, denoted as \(C_0\), is defined by the following formula:

$$C_0 = \frac{\varepsilon_0 A}{d}$$

Here, \(\varepsilon_0\) represents the permittivity of free space.

Effect of Inserting the Dielectric Slab

A dielectric slab is inserted between the plates of the parallel plate capacitor. This slab has a dielectric constant 'K' and a thickness \(t = \frac{3}{5}d\). The area of the slab is the same as the plates, 'A'. When a dielectric slab is placed in this manner, it effectively divides the capacitor into two distinct regions, which can be modeled as two capacitors connected in series.

• The first region is occupied by the dielectric slab.
• The second region is the remaining space, which is an air gap.

Capacitance with Dielectric Slab

The thickness of the dielectric slab is \(t = \frac{3}{5}d\). The capacitance of the part of the capacitor containing this dielectric slab, which we will call \(C_1\), is given by the formula:

$$C_1 = \frac{K \varepsilon_0 A}{t}$$

By substituting the given thickness \(t = \frac{3}{5}d\), we get:

$$C_1 = \frac{K \varepsilon_0 A}{\left(\frac{3}{5}d\right)} = \frac{5K \varepsilon_0 A}{3d}$$

Capacitance with Air Gap

The remaining space between the plates, not occupied by the dielectric slab, forms an air gap. The thickness of this air gap, \(d_{air}\), is:

$$d_{air} = d - t = d - \frac{3}{5}d = \frac{2}{5}d$$

The capacitance of this air gap part, denoted as \(C_2\), is calculated using the formula for a capacitor with air as the dielectric:

$$C_2 = \frac{\varepsilon_0 A}{d_{air}}$$

Substituting the thickness of the air gap \(d_{air} = \frac{2}{5}d\):

$$C_2 = \frac{\varepsilon_0 A}{\left(\frac{2}{5}d\right)} = \frac{5 \varepsilon_0 A}{2d}$$

Calculating the New Capacitance (C') of the Capacitor System

Since the two effective capacitors (one with dielectric and one with air) are connected in series, the total new capacitance \(C'\) (or \(C_{new}\)) of the entire system is determined using the formula for capacitors in series:

$$\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2}$$

Now, substitute the expressions derived for \(C_1\) and \(C_2\) into this equation:

$$\frac{1}{C'} = \frac{3d}{5K \varepsilon_0 A} + \frac{2d}{5 \varepsilon_0 A}$$

To add these fractions, we find a common denominator, which is \(5K \varepsilon_0 A\):

$$\frac{1}{C'} = \frac{3d}{5K \varepsilon_0 A} + \frac{2dK}{5K \varepsilon_0 A}$$

Combine the numerators:

$$\frac{1}{C'} = \frac{3d + 2dK}{5K \varepsilon_0 A}$$

Factor out 'd' from the numerator:

$$\frac{1}{C'} = \frac{d(3 + 2K)}{5K \varepsilon_0 A}$$

To find \(C'\), invert both sides of the equation:

$$C' = \frac{5K \varepsilon_0 A}{d(3 + 2K)}$$

Finally, we want to express \(C'\) in terms of the original capacitance \(C_0 = \frac{\varepsilon_0 A}{d}\). We can rewrite the expression for \(C'\) as:

$$C' = \left(\frac{5K}{3 + 2K}\right) \left(\frac{\varepsilon_0 A}{d}\right)$$

Substituting \(C_0\):

$$C' = C_0 \left(\frac{5K}{2K+3}\right)$$

This expression gives the new capacitance of the parallel plate capacitor after the dielectric slab is inserted. The problem asks for "change in capacitance (C')", but the options clearly indicate the new total capacitance. Therefore, we interpret C' as the new total capacitance.

Conclusion

The new capacitance \(C'\) of the parallel plate capacitor, after inserting a dielectric slab of thickness \(\frac{3}{5}d\) and dielectric constant K, is found to be \(C_0 \left(\frac{5K}{2K+3}\right)\).

This result matches option 3 provided in the question.