In a parallel plate air capacitor of plate separation \(d\), a dielectric slab of thickness \(t\) is introduced between the plates (\(t < d\)). The capacitance becomes one-third of the original value. The dielectric constant of the slab will be:

In a parallel plate air capacitor of plate separation \(d\), a dielectric slab of thickness \(t\) is introduced between the plates (\(t < d\)). The capacitance becomes one-third of the original value. The dielectric constant of the slab will be:

1. \( \frac{t}{d+t} \)

2. \( \frac{t}{2d+t} \)

3. \( \frac{2t}{2d-t} \)

4. \( \frac{t}{d-2t} \)

Solution:

The original capacitance is \( C_0 = \frac{\epsilon_0 A}{d} \). With the dielectric slab inserted, the effective capacitance is given by \[ C = \frac{\epsilon_0 A}{(d-t) + \frac{t}{\kappa}}, \] where \(\kappa\) is the dielectric constant. Since \( C = \frac{1}{3} C_0 \), we have \[ \frac{\epsilon_0 A}{(d-t) + \frac{t}{\kappa}} = \frac{1}{3} \frac{\epsilon_0 A}{d}. \] Canceling common factors yields \[ (d-t) + \frac{t}{\kappa} = 3d. \] Solving for \(\kappa\): \[ \frac{t}{\kappa} = 3d - (d-t) = 2d+t, \] which gives \[ \kappa = \frac{t}{2d+t}. \] Thus, option 2 is the correct answer.