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?

$ϵ_{0} A {\frac{K_{1}}{t_{1}} + \frac{K_{2}}{t_{2}}}$
$C = (K_{1} + K_{2}) ε_{0} A / (t_{1} k_{1} + t_{2} k_{2})$
$C = ε_{0} A / [(t_{1} / K_{1}) + (t_{2} / k_{2})]$
$(K_{1} - K_{2}) ε_{0} \frac{A}{(k_{1} + K_{2})} \cdot (t_{1} + t_{2})$

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?

$ϵ_{0} A {\frac{K_{1}}{t_{1}} + \frac{K_{2}}{t_{2}}}$
$C = (K_{1} + K_{2}) ε_{0} A / (t_{1} k_{1} + t_{2} k_{2})$
$C = ε_{0} A / [(t_{1} / K_{1}) + (t_{2} / k_{2})]$
$(K_{1} - K_{2}) ε_{0} \frac{A}{(k_{1} + K_{2})} \cdot (t_{1} + t_{2})$

Toppr Admin · Accepted Answer

The correct option is A -x3F5-0A-K1t1-K2t2-xA0-We know that the capacitance of the dielectric medium- when introduced between capacitor- is given by C-KE0Ad where A is the cross-sectional area- d is the thickness and k is the dielectric constant-Now Capacity of 1st medium C1-K1E0At1 and-xA0-Capacity of 2nd medium-xA0-C2-K2E0At2Now as the dielectric medium are connected in series thus the total capacity is given as C1-C2-K1E0At1-K2E0At2-E0A-K1t1-K2t2-xA0-

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?ϵ0A{K1t1+K2t2} C=(K1+K2)ε0A/(t1k1+t2k2)C=ε0A/[(t1/K1)+(t2/k2)](K1−K2)ε0A(k1+K2)⋅(t1+t2)