Question 323: Electrostatics & Capacitance

Question

Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance
d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and
dielectric constants K
1
, K
2
and K
3
. The first capacitor is filled as shown in fig.I, and the second one is filled
as shown in fig II.
If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the
two, would be (E
1
refers to capacitor (I) and E
2
to capacitor (II)):

Accepted Answer

$${1 \over {{C_1}}} = {d \over {3A{\varepsilon _0}}}\left( {{1 \over {{K_1}}} + {1 \over {{K_2}}} + {1 \over {{K_3}}}} ight)$$
$${C_1} = {{3A{\varepsilon _0}\left( {{K_1}{K_2}{K_3}} ight)} \over {d\left( {{K_1}{K_2} + {K_2}{K_3} + {K_3}{K_1}} ight)}}$$
$${C_2} = {{A{\varepsilon _0}} \over {3d}}\left( {{K_1} + {K_2} + {K_3}} ight)$$
$${{{E_1}} \over {{E_2}}} = {{{C_1}} \over {{C_2}}} = {{3{K_1}{K_2}{K_3}} \over {\left( {{K_1}{K_2} + {K_2}{K_3} + {K_3}{K_1}} ight)}} 	imes {3 \over {\left( {{K_1} + {K_2} + {K_3}} ight)}}$$
$$ \Rightarrow {{{E_1}} \over {{E_2}}} = {{9{K_1}{K_2}{K_3}} \over {\left( {{K_1} + {K_2} + {K_3}} ight)\left( {{K_1}{K_2} + {K_2}{K_3} + {K_3}{K_1}} ight)}}$$

Answer

$${{{E_1}} \over {{E_2}}} = {{\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)} \over {{K_1}{K_2}{K_3}}}$$

Answer

$${{{E_1}} \over {{E_2}}} = {{{K_1}{K_2}{K_3}} \over {\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}$$

Answer

$${{{E_1}} \over {{E_2}}} = {{\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)} \over {9{K_1}{K_2}{K_3}}}$$

Answer

$${{{E_1}} \over {{E_2}}} = {{9{K_1}{K_2}{K_3}} \over {\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}$$

Electrostatics & Capacitance

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