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Question #42773

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor has plates of area $A$ separated by a distance $d$ . The space between the plates is completely filled with a non-uniform dielectric material. The relative permittivity (dielectric constant) of the material varies linearly with the distance $x$ from the positive plate as $K(x) = K_0(1 + \alpha x)$ , where $K_0$ is a dimensionless positive constant and $\alpha$ is a positive constant with dimensions of inverse length. If $\varepsilon_0$ is the permittivity of free space, what is the capacitance of this capacitor?

Options

A.
$\frac{\varepsilon_0 A K_0 \alpha}{\ln(1 + \alpha d)}$
B.
$\frac{\varepsilon_0 A K_0 \left(1 + \frac{\alpha d}{2}\right)}{d}$
C.
$\frac{\varepsilon_0 A K_0}{d \ln(1 + \alpha d)}$
D.
$\frac{\varepsilon_0 A K_0}{\alpha d^2} \ln(1 + \alpha d)$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Dielectric Polarization Logic|Electric Field Symmetry|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the conceptual shift from using standard permittivity formulas to applying fundamental Gauss's Law principles in non-uniform media.

Students often fall into the trap of using the standard formula $C = \frac{\varepsilon_0 K A}{d}$ by either taking an arithmetic mean of the dielectric constant $K_{\text{avg}} = K_0(1 + \alpha d/2)$ (Option B) or by incorrectly integrating the capacitance formula directly.

The core misconception is failing to recognize that in a series configuration (which a varying $K(x)$ along the field lines represents), it is the Electric Displacement field $\vec{D}$ (and thus the free charge density $\sigma$ ) that remains constant, not the Electric Field $\vec{E}$ .

By applying Gauss's Law for dielectrics ( $\nabla \cdot \vec{D} = \rho_{\text{free}}$ ), one correctly identifies that $E(x)$ must vary inversely with $K(x)$ , requiring a spatial integration of $E(x)$ to find the true potential difference.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Dielectric Polarization Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free