Question

The capacitance of a parallel plate capacitor is 2.5 µF. When it is half filled with a dielectric as shown in figure, its capacitance becomes 5 μF. The dielectric constant of the dielectric is

• A. 7.5
• B. 3
• C. 4
• D. 5

Hint:

Logic Tip: When capacitors are in parallel, their equivalent capacitance is the sum of individual values.

Answer 1

The Correct Option is B

Concept: When a dielectric slab is inserted in part of the area of a parallel plate capacitor, the system behaves like two capacitors connected in parallel, since both parts have the same potential difference.
Step 1: Initial capacitance \[ C = \frac{\varepsilon_0 A}{d} = 2.5\,\mu\text{F} \]
Step 2: Divide into two regions The plate is divided into two equal areas $\frac{A}{2}$:

• Without dielectric: \[ C_1 = \frac{\varepsilon_0 (A/2)}{d} = \frac{C}{2} = 1.25\,\mu\text{F} \]
• With dielectric: \[ C_2 = \frac{\varepsilon_r \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_r C}{2} = 1.25\varepsilon_r\,\mu\text{F} \]

Step 3: Equivalent capacitance Since they are in parallel: \[ C_{\text{eq = C_1 + C_2 \] \[ 5 = 1.25 + 1.25\varepsilon_r \]
Step 4: Solve for $\varepsilon_r$ \[ 1.25\varepsilon_r = 5 - 1.25 = 3.75 \] \[ \varepsilon_r = \frac{3.75}{1.25} = 3 \] Final Answer: \[ \boxed{\varepsilon_r = 3} \]