Question

Find the capacitance between the plates shown in the figure below

• A. $\frac{{\in }_0{K}_1{K}_{2}bl}{d(K_2-K_1)}ln\frac{K_2}{K_1}$
• B. $\frac{{\in }_0{K}_1{K}_{2}bl}{2d(K_2-K_1)}ln\frac{K_2}{K_1}$
• C. $\frac{{\in }_0{K}_1{K}_{2}bl}{3d(K_2-K_1)}ln\frac{K_2}{K_1}$
• D. None of the above

Answer 1

The correct option is A $\frac{{\in }_0{K}_1{K}_{2}bl}{d(K_2-K_1)}ln\frac{K_2}{K_1}$

For this type of problem, we will divide this into infinitely small parts and will apply integration.
These $K_1$ and $K_2$ are separated by slant but as the $dy$ is very small we can consider is straight.
Let $d{C}_1$ be the capacitance of capacitor having $K_1$ and height $dy$
$d{C}_1=\frac{K_1{\in }_0\left(bdy\right)}{x}$
$d{C}_1=\frac{K_2{\in }_0\left(bdy\right)}{d-x}$
$dC=\frac{\frac{K_1{\in }_0\left(bdy\right)}{x}.\frac{K_2{\in }_0\left(bdy\right)}{d-x}}{\frac{K_1{\in }_0\left(bdy\right)}{x}+\frac{K_2{\in }_0\left(bdy\right)}{d-x}}$
$=\frac{\frac{K_1{K}_2{\in }_{0}bdy}{x(d-x)}}{\frac{K_1}{x}+\frac{K_2}{d-x}}$
$=\frac{\frac{K_1{K}_2{\in }_{0}bdy}{x(d-x)}}{\frac{K_1(d-x)+K_2\left(x\right)}{x(d-x)}}$
$dC=\frac{K_1{K}_2{\in }_{0}bdy}{K_1(d-x)+K_2\left(x\right)}$
To integrate above equation we have to change this unknown variable $x$ in terms of known quantities.
So, by the property of similar triangle
$\frac{x}{y}=\frac{d}{l}$

So, $x=\frac{d.y}{l}$

$dC=\frac{K_1{K}_2{\in }_{0}bdy}{K_1(d-\frac{d.y}{l})+K_2\left(\frac{d.y}{l}\right)}$
$dC=\frac{K_1{K}_2{\in }_{0}bl}{d}\left[\frac{dy}{K_1(l-y)+K_{2}y}\right]$

$dC=\frac{K_1{K}_2{\in }_{0}bl}{d}\left[\frac{dy}{K_{1}l-K_{1}y+K_{2}y}\right]$

$dC=\frac{K_1{K}_2{\in }_{0}bl}{d}\left[\frac{dy}{K_{1}l+(K_2-K_1)y}\right]$

These capacitor can be considered as connected to a point as shown in figure there are parallel with each other.

so now we will do integration to get total capacitance.
$C=\frac{K_1{K}_2{\in }_{0}bl}{d}ln\frac{K_{1}l+(K_2-K_1)y}{K_2-K_1}{|}_0^l$
$C=\frac{K_1{K}_2{\in }_{0}bl}{d(K_2-K_1)}\left[ln\right(K_{2}l)-ln(K_{1}l\left)\right]$
$C=\frac{{\in }_0{K}_1{K}_{2}bl}{d(K_2-K_1)}ln\frac{K_2}{K_1}$