Question

The gap between the plates of a parallel plate capacitor of area 'A' & distance between plates 'd' is filled with a dielectric whose permitivity varies linearly from 'k1' at one plate to 'k2' at other. find the capacitance of the capacitor.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{C=\frac{A \epsilon_{o} ( k_{2} - k_{1})}{ln \bigg(\frac{ k_{2} }{ k_{1} } \bigg) d}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \bold {\underline{Given :}}}$

$\tt: \implies Area \: of \: plate = A$

$\tt: \implies Seperation \: between \: plates = d$

$\red{ \bold {\underline{To \: Find:}}}$

$\tt: \implies Capacitance \: of \: capacitor =?$

• According to given question :

$\bold{As \: we \: know \: that}$

$\tt: \implies tan \: \theta = m = \frac{ k_{2} - k_{1} }{d} - - - - - (1)$

$\bold{As \: we \: know \: that}$

$\tt: \implies Eqn \: of \: line \to y = mx + c$

$\tt: \implies k = \bigg(\frac{ k_{2} - k_{1} }{d} \bigg)x + k_{1} - - - - - (2)$

$\bold{As \: we \: know \: that} \\ \tt: \implies dC = \frac{kA \epsilon_{o} }{dx}$

$\tt: \implies dC = \frac{A \epsilon_{o} }{dx} \bigg( \frac{( k_{2} - k_{1})x}{d} + k_{1} \bigg)$

$\bold{For \: series \: capacitance:}$

$\tt: \implies \frac{1}{dC} = \frac{1}{A \epsilon_{o} } \bigg( \frac{dx}{ \frac{ (k_{2} - k_{1})x }{d} + k_{1} } \bigg)$

$\tt: \implies \frac{1}{C}=\int \frac{1}{dC} = \frac{1}{A \epsilon _{o} } \int \limits_{0}^{d} \bigg( \frac{dx}{ \frac{ ( k_{2} - k_{1})x}{d} + k_{1}} \bigg)$

$\tt \circ \: \int \frac{dx}{bx + a} = \frac{ln(bx + a)}{b} + c$

$\tt: \implies \frac{1}{C} = \frac{1}{A\epsilon_{o} } \bigg( \frac{ln \frac{( k_{2} - k_{1} ) {x} }{d} + k_{1}}{ \frac{ k_{2} - k_{1}}{d} } \bigg)_{0}^{d}$

$\tt: \implies \frac{1}{C} = \frac{d}{A \epsilon_{o}( k_{2} - k_{1}) } \bigg(ln \frac{ (k_{2} - k_{1} )x}{d} + k_{1} \bigg)_{0}^{d}$

$\tt: \implies \frac{1}{C} = \frac{d}{A\epsilon_{o}( k_{2} - k_{1} ) } \bigg(ln \frac{ (k_{2} - k_{1} )d}{d} + k_{1} - ln (k_{1}) \bigg)$

$\tt: \implies \frac{1}{C} =\frac{d}{A \epsilon_{o}( k_{2} - k_{1} ) } (ln( k_{2} )- ln(k_{1}))$

$\tt: \implies \frac{1}{C} =\frac{d}{A \epsilon_{o}( k_{2} - k_{1} ) }ln \bigg( \frac{ k_{2} }{ k_{1}} \bigg)$

$\green{\tt: \implies C = \frac{A \epsilon_{o} ( k_{2} - k_{1})}{ln \bigg(\frac{ k_{2} }{ k_{1} } \bigg) d} }$

Answer 2

Assume there exists $\displaystyle\sf {n}$ divisions of same thickness in between the plates of the capacitor and each division be filled with dielectric of permittivity each varying linearly from $\displaystyle\sf {K_1}$ and $\displaystyle\sf {K_2.}$

The permittivity of dielectric in first division,

• $\displaystyle\sf {K_i=K_1}$

The permittivity of dielectric in second division (say),

• $\displaystyle\sf {K_{i+\Delta i}=K_1+\Delta K}$

Likewise, the permittivity of dielectric in third division,

• $\displaystyle\sf {K_{i+2\Delta i}=K_1+2\Delta K}$

Generally, the permittivity of dielectric in $\displaystyle\sf {r^{th}}$ division,

• $\displaystyle\sf {K_{i+r\Delta i}=K_1+r\Delta K}$

So, the permittivity of dielectric in last division, i.e., $\displaystyle\sf {K_2,}$ is given by,

$\displaystyle\sf{\longrightarrow K_2=K_1+n\Delta K}$

$\displaystyle\sf{\longrightarrow \Delta K=\dfrac{K_2-K_1}{n}\quad\quad\dots(1)}$

The $\displaystyle\sf {n}$ divisions are of same thickness, so since the distance between the plates is equal to $\displaystyle\sf {d,}$ each thickness measures $\displaystyle\sf {\dfrac {d}{n}.}$

Each division is in series, so the capacitance of the capacitor will be,

$\displaystyle\sf{\longrightarrow C=\dfrac {A\epsilon_0}{\dfrac {d}{n}\left [\dfrac {1}{K_i}+\dfrac {1}{K_{i+\Delta i}}+\dfrac {1}{K_{i+2\Delta i}}+\,\dots\,+\dfrac {1}{K_{i+n\Delta i}}\right]}}$

$\displaystyle\sf{\longrightarrow C=\dfrac {A\epsilon_0}{\dfrac {d}{n}\left [\dfrac {1}{K_1}+\dfrac {1}{K_1+\Delta K}+\dfrac {1}{K_1+2\Delta K}+\,\dots\,+\dfrac {1}{K_2}\right]}}$

Or,

$\displaystyle\sf{\longrightarrow C=\dfrac {A\epsilon_0}{d\displaystyle\sf{\sum_{r=0}^n\dfrac {1}{K_1+r\Delta K}\cdot\dfrac {1}{n}}}}$

As the permittivity of dielectric in each division varies linearly, there should be variation in infinitely many divisions between the plates of infinitesimal thickness.

Hence effective capacitance of the capacitor is actually,

$\displaystyle\sf{\longrightarrow C=\dfrac {A\epsilon_0}{d\displaystyle\sf{\lim_{n\to\infty}\sum_{r=0}^n\dfrac {1}{K_1+r\Delta K}\cdot\dfrac {1}{n}}}}$

Multiplying both numerator and denominator by $\displaystyle\sf {K_2-K_1,}$

$\displaystyle\sf{\longrightarrow C=\dfrac {A\epsilon_0(K_2-K_1)}{d\displaystyle\sf{\lim_{n\to\infty}\sum_{r=0}^n\dfrac {1}{K_1+r\Delta K}\cdot\dfrac {K_2-K_1}{n}}}}$

From (1),

$\displaystyle\sf{\longrightarrow C=\dfrac {A\epsilon_0(K_2-K_1)}{d\displaystyle\sf{\lim_{n\to\infty}\sum_{r=0}^n\dfrac {1}{K_1+r\Delta K}\cdot\Delta K}}\quad\quad\dots(2)}$

Now the sum in the denominator is a Riemann Sum which can be converted to a definite integral as,

$\displaystyle\boxed {\sf {\lim_{n\to\infty}\sum_{r=0}^nf(a+r\Delta x)\cdot\Delta x=\int\limits_a^bf(x)\ dx\ where\ \Delta x=\dfrac {b-a}{n}}}$

Thus (2) becomes,

$\displaystyle\mathsf{\longrightarrow C=\dfrac {A\epsilon_0(K_2-K_1)}{d\displaystyle\mathsf{\int\limits_{K_1}^{K_2}\dfrac {1}{x}\ dx}}}$

$\displaystyle\sf {\longrightarrow\underline {\underline {C=\dfrac {A\epsilon_0(K_2-K_1)}{d\log\left (\dfrac {K_2}{K_1}\right)}}}}$