Question

If half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant 4, the capacitance is \( C_1 \). If one-third of the space between the plates of the capacitor is filled with the medium of dielectric constant 4, the capacitance is \( C_2 \). If in both cases, the dielectric is placed parallel to the plates of the capacitor, then \( C_1 : C_2 \) is:

• A. \( 2:3 \)
• B. \( 4:3 \)
• C. \( 6:5 \)
• D. \( 7:5 \)

Hint:

For partially filled capacitors, use the series formula for dielectric sections.
- The total capacitance depends on the fraction of dielectric and its dielectric constant.

Answer 1

The Correct Option is C

For a parallel plate capacitor partially filled with a dielectric, the equivalent capacitance is given by: \[ C = \frac{\varepsilon_0 A}{d} \frac{1}{\frac{x}{\kappa} + (1-x)} \] where: - \( x \) is the fraction of the dielectric-filled space, - \( \kappa \) is the dielectric constant. 1. Case 1: Half-filled (\( x = \frac{1}{2} \), \( \kappa = 4 \)) \[ \frac{1}{C_1} = \frac{1}{C_0} \left(\frac{1}{\frac{1}{2 \times 4} + \frac{1}{2}} \right) \] \[ \frac{1}{C_1} = \frac{1}{C_0} \left(\frac{1}{\frac{1}{8} + \frac{1}{2}} \right) \] \[ C_1 = \frac{8}{5} C_0 \] 2. Case 2: One-third filled (\( x = \frac{1}{3} \), \( \kappa = 4 \)) \[ \frac{1}{C_2} = \frac{1}{C_0} \left(\frac{1}{\frac{1}{3 \times 4} + \frac{2}{3}} \right) \] \[ \frac{1}{C_2} = \frac{1}{C_0} \left(\frac{1}{\frac{1}{12} + \frac{2}{3}} \right) \] \[ C_2 = \frac{6}{5} C_0 \] 3. Ratio of \( C_1 : C_2 \): \[ \frac{C_1}{C_2} = \frac{8}{5} C_0 : \frac{6}{5} C_0 = 6:5 \] Thus, the correct answer is \(\boxed{6:5}\).