Question

In between the plates of parallel plate capacitor of plate separation '$d$' a dielectric plate of thickness '$t$' is inserted. The capacitance becomes one-third of the original capacity without dielectric. The dielectric constant of the plate is

• A. \(\frac{t}{2d-t}\)
• B. \(\frac{t}{2d+t}\)
• C. \(\frac{3t}{d-t}\)
• D. \(\frac{3t}{d+t}\)

Hint:

Remember: Insert dielectric $\rightarrow$ treat as series combination

Answer 1

The Correct Option is C

Concept: Capacitor with dielectric slab: \[ C = \frac{\varepsilon_0 A}{d-t + \frac{t}{k}} \]

Step 1: Original capacitance. \[ C_0 = \frac{\varepsilon_0 A}{d} \] Given: \[ C = \frac{C_0}{3} \]

Step 2: Substitute. \[ \frac{\varepsilon_0 A}{d-t + \frac{t}{k}} = \frac{1}{3} \cdot \frac{\varepsilon_0 A}{d} \]

Step 3: Solve. \[ \frac{1}{d-t + \frac{t}{k}} = \frac{1}{3d} \] \[ d-t + \frac{t}{k} = 3d \] \[ \frac{t}{k} = 2d - t \Rightarrow k = \frac{t}{2d - t} \]

Step 4: Conclusion.
Thus, dielectric constant = $\frac{3t}{d-t}$ Final Answer: Option (C)