Question 459: Electrostatics & Capacitance

Question

The plates of a parallel plate capacitor are separated by a distance 'd' with air as the medium between them. A dielectric slab of dielectric constant 3 is introduced between the plates so as to increase the capacity by $50 \%$. The thickness of the dielectric slab is

Accepted Answer

Step 1: Calculate the initial capacitance of the parallel plate capacitor.
When the medium between the plates is air (or vacuum), the capacitance ($C_0$) is given by:
$$C_0 = \frac{\epsilon_0 A}{d}$$
where:
$\epsilon_0$ is the permittivity of free space.
$A$ is the area of each plate.
$d$ is the separation between the plates.
Step 2: Calculate the capacitance after introducing the dielectric slab.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced between the plates of a parallel plate capacitor, the new capacitance ($C_f$) is given by the formula:
$$C_f = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$$
In this problem, the dielectric constant $K = 3$. So, substituting this value:
$$C_f = \frac{\epsilon_0 A}{d - t + \frac{t}{3}}$$
To simplify the denominator:
$$C_f = \frac{\epsilon_0 A}{d - \frac{3t}{3} + \frac{t}{3}}$$
$$C_f = \frac{\epsilon_0 A}{d - \frac{2t}{3}}$$
Step 3: Use the given condition about the increase in capacity.
The problem states that the capacity increases by $50 \%$. This means the final capacitance $C_f$ is $150\%$ of the initial capacitance $C_0$.
$$C_f = C_0 + 50\% 	ext{ of } C_0$$
$$C_f = C_0 + 0.5 C_0$$
$$C_f = 1.5 C_0$$
Step 4: Equate the expressions for $C_f$ and solve for the thickness $t$.
Now, substitute the expressions for $C_f$ and $C_0$ into the equation from Step 3:
$$\frac{\epsilon_0 A}{d - \frac{2t}{3}} = 1.5 \left( \frac{\epsilon_0 A}{d} ight)$$
We can cancel out $\epsilon_0 A$ from both sides of the equation:
$$\frac{1}{d - \frac{2t}{3}} = \frac{1.5}{d}$$
Now, cross-multiply to solve for $t$:
$$d = 1.5 \left( d - \frac{2t}{3} ight)$$
$$d = 1.5d - 1.5 	imes \frac{2t}{3}$$
$$d = 1.5d - \frac{3}{2} 	imes \frac{2t}{3}$$
$$d = 1.5d - t$$
Rearrange the equation to isolate $t$:
$$t = 1.5d - d$$
$$t = 0.5d$$
$$t = \frac{d}{2}$$
Thus, the thickness of the dielectric slab is $\frac{d}{2}$.
The final answer is $\boxed{\frac{\mathrm{d}}{2}}$.

Answer

$\frac{\mathrm{d}}{2}$

Answer

$\frac{\mathrm{d}}{3}$

Answer

$\frac{\mathrm{d}}{5}$

Answer

$\frac{5 \mathrm{~d}}{6}$

Electrostatics & Capacitance

The plates of a parallel plate capacitor are separated by a distance 'd' with air as the medium between them. A dielectric slab of dielectric constant 3 is introduced between the plates so as to increase the capacity by 50 \%. The thickness of the dielectric slab is

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