A system of two parallel plates- each of area A- are separated by distances d-1 and d-2- The space between them is filled with dielectrics of permittivities epsilon-1 and epsilon-2- The permittivity of space is epsilon-0- The equivalent capacitance of the system is -dfrac -epsilon-1epsilon-2A-epsilon-2d-1-epsilon-1d-2-dfrac -epsilon-1epsilon-2epsilon-0A-epsilon-1d-1-epsilon-2d-2-dfrac -epsilon-0A-epsilon-1d-1-epsilon-2d-2-dfrac -epsilon-0A-epsilon-1d-2-epsilon-2d-1-

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Toppr Admin · Accepted Answer

There are two capacitors C1 and C2 and they are in series-xA0-Here- C1-A-x3F5-1d1 and C2-A-x3F5-2d2-xA0-The equivalent capacitance is Ceq-C1C2C1-C2-A-x3F5-1-x3F5-2-x3F5-1d2-x3F5-2d1

There are two capacitors C1 and C2 and they are in series. Here, C1=Aϵ1d1 and C2=Aϵ2d2 The equivalent capacitance is Ceq=C1C2C1+C2=Aϵ1ϵ2ϵ1d2+ϵ2d1

There are two capacitors $C_{1}$ and $C_{2}$ and they are in series.
Here, $C_{1} = \frac{A ϵ_{1}}{d_{1}}$ and $C_{2} = \frac{A ϵ_{2}}{d_{2}}$
The equivalent capacitance is $C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{A ϵ_{1} ϵ_{2}}{ϵ_{1} d_{2} + ϵ_{2} d_{1}}$