A parallel plate capacitor with plate area $$A$$ and plate separation $$d = 2$$ m has a capacitance of $$4$$ $$\mu$$F. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $$K = 3$$ (as shown in figure) will be

Given:
Plate separation d=2 m, original capacitance C=4 μF, dielectric constant K=3

Step 1: Divide the capacitor into two regions
Each region has thickness d/2, so they act as series capacitors.

Step 2: Capacitance of each part

• Without dielectric:

$$C_1=\frac{\varepsilon_0A}{d/2}=\frac{2\varepsilon_0A}{d}=2C$$

• With dielectric K=3K:

$$C_2=\frac{K\varepsilon_0A}{d/2}=\frac{2K\varepsilon_0A}{d}=2KC=6C$$

Step 3: Equivalent capacitance (series)

$$C_{\text{new}}=\frac{C_1C_2}{C_1+C_2}$$

$$=\frac{(2C)(6C)}{2C+6C}$$

$$=\frac{12C^2}{8C}$$

$$Cnew=\frac{3}{2}C​$$

Step 4: Substitute value

$$C_{new}​=\ \frac{\ 3}{2}​\times4=6μF$$

Final Answer:

6 μF