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Question

The space between two plates of a condenser is filled with two dielectric media of thickness $t_{1}$ and $t_{2}$ and dielectric constant $k_{1}$ and $k_{2}$ respectively. The capacity of the condenser is given by?

ϵ_{0} A {\frac{K_{1}}{t_{1}} + \frac{K_{2}}{t_{2}}}

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C = (K_{1} + K_{2}) ε_{0} A / (t_{1} k_{1} + t_{2} k_{2})

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C = ε_{0} A / [(t_{1} / K_{1}) + (t_{2} / k_{2})]

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(K_{1} - K_{2}) ε_{0} \frac{A}{(k_{1} + K_{2})} \cdot (t_{1} + t_{2})

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Solution

The correct option is A $ϵ_{0} A {\frac{K_{1}}{t_{1}} + \frac{K_{2}}{t_{2}}}$
We know that the capacitance of the dielectric medium, when introduced between capacitor, is given by C $= K \frac{E_{0} A}{d}$ where A is the cross-sectional area, d is the thickness and k is the dielectric constant.
Now Capacity of 1st medium $C_{1} = K_{1} \frac{E_{0} A}{t_{1}}$ and Capacity of 2nd medium $C_{2} = K_{2} \frac{E_{0} A}{t_{2}}$
Now as the dielectric medium are connected in series thus the total capacity is given as $C_{1} + C_{2}$ = $= \frac{K_{1} E_{0} A}{t_{1}} + \frac{K_{2} E_{0} A}{t_{2}} = E_{0} A {\frac{K_{1}}{t_{1}} + \frac{K_{2}}{t_{2}}}$

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The space between two plates of a condenser is filled with two dielectric media of thickness t1 and t2 and dielectric constant k1 and k2 respectively. The capacity of the condenser is given by?

The space between two plates of a condenser is filled with two dielectric media of thickness $t_{1}$ and $t_{2}$ and dielectric constant $k_{1}$ and $k_{2}$ respectively. The capacity of the condenser is given by?