Question 250: Electrostats & Capacitance

Question

A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1 + $$\alpha $$x) where 'x' is the distance measured from one of the plates. If (ad) << 1, the total capacitance of the system is best given by the expression :

Accepted Answer

k(x) = K(1 + $$\alpha $$x)
Capacitance of element dC
1
= $${{K{\varepsilon _0}A} \over {dx}}$$ = $${{K\left( {1 + \alpha x} \right){\varepsilon _0}A} \over {dx}}$$
$${1 \over {{C_{eq}}}}$$ = $$\int {{1 \over {d{C_1}}}} $$
= $$\int\limits_0^d {{{dx} \over {KA\left( {1 + \alpha x} \right){\varepsilon _0}}}} $$
= $${{1 \over {KA\alpha {\varepsilon _0}}}\ln \left( {1 + \alpha x} \right)}$$
Using expansion of ln (1 + x) keeping x $$ \ll $$ 1
$$ \Rightarrow $$ $${1 \over {{C_{eq}}}}$$ = $${{1 \over {KA\alpha {\varepsilon _0}}}\left( {\alpha d - {{{\alpha ^2}{d^2}} \over 2}} \right)}$$
$$ \Rightarrow $$ C
eq
= $${{{{\varepsilon _0}KA} \over d}{{\left( {1 - {{\alpha d} \over 2}} \right)}^{ - 1}}}$$
$$ \Rightarrow $$ C
eq
= $${{{{\varepsilon _0}KA} \over d}\left( {1 + {{\alpha d} \over 2}} \right)}$$

Answer

$${{A{ \in _0}K} \over d}\left( {1 + {{\left( {{{\alpha d} \over 2}} \right)}^2}} \right)$$

Answer

$${{A{ \in _0}K} \over d}\left( {1 + {{\alpha d} \over 2}} \right)$$

Answer

$${{A{ \in _0}K} \over d}\left( {1 + {{{\alpha ^2}{d^2}} \over 2}} \right)$$

Answer

$${{A{ \in _0}K} \over d}\left( {1 + \alpha d} \right)$$

Electrostats & Capacitance

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