Question

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

• A. $\frac{t}{2d+t}$
• B. $\frac{t}{d-2t}$
• C. $\frac{t}{d+t}$
• D. $\frac{2t}{2d-t}$

Answer 1

Answer: (A)

$\frac{t}{2d+t}$

Answer 2

1. Original Capacitance: The capacitance of a parallel plate air capacitor with plate area $A$ and separation $d$ is given by: $C_0 = \frac{\epsilon_0 A}{d}$

2. Capacitance with Dielectric Slab: When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced between the plates, the capacitor can be considered as a series combination of two air gaps (total thickness $d-t$) and one dielectric slab (thickness $t$). The effective capacitance $C$ is given by: $C = \frac{\epsilon_0 A}{(d-t) + \frac{t}{K}}$

3. Using the Given Condition: The problem states that the new capacitance $C$ becomes one-third of the original capacitance $C_0$: $C = \frac{1}{3} C_0$

4. Substituting and Solving: Substitute the expressions for $C$ and $C_0$ into the equation: $\frac{\epsilon_0 A}{(d-t) + \frac{t}{K}} = \frac{1}{3} \frac{\epsilon_0 A}{d}$

   Cancel out $\epsilon_0 A$ from both sides: $\frac{1}{(d-t) + \frac{t}{K}} = \frac{1}{3d}$

   Take the reciprocal of both sides: $(d-t) + \frac{t}{K} = 3d$

   Isolate the term with $K$: $\frac{t}{K} = 3d - (d-t)$ $\frac{t}{K} = 3d - d + t$ $\frac{t}{K} = 2d + t$

   Solve for $K$: $K = \frac{t}{2d+t}$