Question

A parallel plate capacitor has square plates of edge length $e$ and plate separation $d$. The gap between the plates is filled with a dielectric of dielectric constant $K$ which varies parallel to an edge as ${K=K}_0+\alpha x$. Here $x$ is the distance along the edge of the capacitor.


where $K_0$ and $\alpha$ are constants. The capacitance of system will be :

• A. $3{\epsilon }_0{e}^2(K_0+\frac{\alpha e}{2})/d$
• B.
  $2{\epsilon }_0{e}^2(K_0+\frac{\alpha e}{2})/d$
• C.
  ${\epsilon }_0{e}^2(K_0+\frac{\alpha e}{2})/d$
• D.
  ${\epsilon }_0{e}^2(K_0+\frac{\alpha e}{2})/2d$

Answer 1

The correct option is C

${\epsilon }_0{e}^2(K_0+\frac{\alpha e}{2})/d$
Let choose a small element of thickness $dx$ at a distance $x$ from the left end as shown in the figure.


$\text{Area of this small capacitor is,}dA=edx$

Dielectric constant at this location,

$K=K_0+\alpha x$

The capacitance of small capacitor is,

$dC=\frac{(K_0+\alpha x){\epsilon }_{0}edx}{d}$

Now all such capacitors will be in parallel connection because their two ends are connected at same potential difference.

Thus using the integration technique to find the summation of all such capacitors we get,

$C=\int \text{dC}={\int }_0^e\frac{(K_0+\alpha x){\epsilon }_{0}edx}{d}$

$C=\frac{{\epsilon }_{0}e}{d}[K_0\text{e}+\frac{\alpha e^2}{2}]$

$\Rightarrow C=\frac{{\epsilon }_0{\text{e}}^2}{d}[K_0+\frac{\alpha \text{e}}{2}]$

Why this question ? $\underset{–}{\text{Tip}}:$In this problem we can imagine the whole capacitor as made up of a large number of small strips, which will serve as a single capacitor & arranged in parallel connection.