Question 637: Electrostatics & Capacitance

Question

A parallel-plate capacitor of area a, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k
1
, k
2
, k
3
and k
4
as shown in the figure. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

Accepted Answer

Here, $${C_1} = {{2{\varepsilon _0}{k_1}A} \over {3d}},{C_2} = {{2{\varepsilon _0}{k_2}A} \over {3d}}$$
$${C_3} = {{2{\varepsilon _0}{k_3}A} \over {3d}},{C_4} = {{2{\varepsilon _0}{k_4}A} \over d}$$
Given system of C
1
, C
2
, C
3
and C
4
can be simplified as
$$ 	herefore $$ $${1 \over {{C_{AB}}}} = {1 \over {{C_1} + {C_2} + {C_3}}} + {1 \over {{C_4}}}$$
Suppose, $${C_{AB}} = {{k{\varepsilon _0}A} \over d}$$
$${C_{AB}} = {1 \over {k\left( {{{{\varepsilon _0}A} \over d}} ight)}} = {1 \over {{2 \over 3}{{{\varepsilon _0}A} \over d}\left( {{k_1} + {k_2} + {k_3}} ight)}} + {1 \over {{{2{\varepsilon _0}A} \over d}{k_4}}}$$
$$ \Rightarrow {1 \over k} = {1 \over {2\left( {{k_1} + {k_2} + {k_3}} ight)}} + {1 \over {2{k_4}}}$$
$$ 	herefore {2 \over k} = {3 \over {{k_1} + {k_2} + {k_3}}} + {1 \over {{k_4}}}$$

Answer

k = k
1
+ k
2
+ k
3
+ 3k
4

Answer

k = $${2 \over 3}$$ (k
1
+ k
2
+ k
3
) + 2k
4

Answer

$${2 \over k}$$ = $${3 \over {{k_1} + {k_2} + {k_3}}} + {1 \over {{k_4}}}$$

Answer

$${1 \over k}$$ = $${1 \over {{k_1}}} + {1 \over {{k_2}}} + {1 \over {{k_3}}} + {3 \over {2{k_4}}}$$

Electrostatics & Capacitance

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