In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be: (Given the area of the plate = $$A$$)

We need to determine the expression for the equivalent capacitance of a parallel plate capacitor filled with three different dielectric slabs in a compound arrangement.

1. Identify individual Capacitances

From the diagram , the three dielectrics are placed back-to-back across the plate separation distance. This layout forms a series combination of three separate capacitors ($$C_1$$, $$C_2$$, and $$C_3$$):

• First Section ($$C_1$$):
  Dielectric constant = $$K$$, thickness = $$d$$

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

• Second Section ($$C_2$$):
  Dielectric constant = $$3K$$, thickness = $$2d$$

  $$C_2 = \frac{3K \varepsilon_0 A}{2d}$$

• Third Section ($$C_3$$):
  Dielectric constant = $$5K$$, thickness = $$3d$$

  $$C_3 = \frac{5K \varepsilon_0 A}{3d}$$

2. Apply the Series Combination Formula

For capacitors connected in series, the equivalent capacitance ($$C_{\text{eq}}$$) is given by:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

Substitute the reciprocal values of $$C_1$$, $$C_2$$, and $$C_3$$ into the expression:

$$\frac{1}{C_{\text{eq}}} = \frac{d}{K \varepsilon_0 A} + \frac{2d}{3K \varepsilon_0 A} + \frac{3d}{5K \varepsilon_0 A}$$

Factor out the common term $$\frac{d}{K \varepsilon_0 A}$$:

$$\frac{1}{C_{\text{eq}}} = \frac{d}{K \varepsilon_0 A} \left( 1 + \frac{2}{3} + \frac{3}{5} \right)$$

3. Simplify the Fraction and Double-Check the Calculation

Find a common denominator for the terms inside the parentheses (which is $$15$$):

$$1 + \frac{2}{3} + \frac{3}{5} = \frac{15}{15} + \frac{10}{15} + \frac{9}{15} = \frac{15 + 10 + 9}{15} = \frac{34}{15}$$

Now substitute this back into our equation for $$\frac{1}{C_{\text{eq}}}$$:

$$\frac{1}{C_{\text{eq}}} = \frac{d}{K \varepsilon_0 A} \left( \frac{34}{15} \right) = \frac{34d}{15K \varepsilon_0 A}$$

Inverting the expression gives the final total capacity:

$$C_{\text{eq}} = \frac{15}{34} \frac{K \varepsilon_0 A}{d}$$

Conclusion

The total capacity expression matches Option A perfectly:

$$C_{\text{eq}} = \frac{15}{34} \frac{K \varepsilon_0 A}{d}$$