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A parallel plate capacitor is filled with two layers of different materials A and B as shown in the figure. Material A has dielectric constant $ε_{1}$ and conductivity $σ_{1}$ , while material B has dielectric constant $ε_{2}$ and conductivity $σ_{2}$ . The capacitor is connected across a battery of terminal voltage $V$ .

Electric field in material A is

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Answer:

Solution:

$R_{1} = \frac{d _{1}}{σ _{1} A}$ , and $R_{2} = \frac{d _{2}}{σ _{2} A}$ $R_{e q} = \frac{1}{A} [\frac{d _{1} σ _{2} + d _{2} σ _{1}}{σ _{1} σ _{2}}]$

$I = \frac{V}{R _{eq}} = \frac{σ _{1} σ _{2} AV}{d _{1} σ _{2} + d _{2} σ _{1}} \Rightarrow E_{A} = \frac{R _{1} I}{d _{1}} = \frac{d _{1}}{σ _{1} Ad _{1}} \times \frac{σ _{1} σ _{2} AV}{d _{1} σ _{2} + d _{2} σ _{1}} = \frac{σ _{2} V}{d _{1} σ _{2} + d _{2} σ _{1}}$

Similarly $E_{B} = \frac{σ _{1} V}{d _{1} σ _{2} + d _{2} σ _{1}} \Rightarrow \frac{σ A}{ε _{0}} = ϕ = A (E_{B} - E_{A}) \Rightarrow σ = ε_{0} (E_{B} - E_{A}) = \frac{ε _{0} V ( σ _{1} - σ _{2} )}{d σ _{2} + d _{2} σ _{1}}$

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