Question

When a parallel plate capacitor is charged up to 95 V, its capacitance is \( C \). If a dielectric slab of thickness 2 mm is inserted between plates and the plate separation is increased by 1.6 mm such that the potential difference remains constant, find the dielectric constant of the material:

• A. \( 2.4 \)
• B. \( 4.5 \)
• C. \( 5.0 \)
• D. \( 9.0 \)

Hint:

The capacitance of a capacitor with a dielectric slab depends on the thickness of the slab and its dielectric constant. The new capacitance is calculated using modified plate separation.

Answer 1

The Correct Option is C

Step 1: Understanding the Capacitance Relation The new capacitance with a dielectric slab is given by: \[ \frac{d - t + \frac{t}{K}}{d} = \frac{1}{K} \] where: \( d' = d + 1.6 \) mm, \( t = 2 \) mm (thickness of dielectric slab), \( K \) is the dielectric constant. Step 2: Solve for \( K \) \[ \frac{(d+1.6) - 2 + \frac{2}{K}}{d+1.6} = \frac{1}{K} \] Solving for \( K \), we get: \[ K = 5.0 \] Thus, the correct answer is \( 5.0 \).