1
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1
A parallel plate capacitor with square plates is filled with four dielecytrics of dielectrics constants K1, K2, K3, K4 arranged as shown in the figure. The effective dielectric constant K will be :

A
$$K = {{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}$$
B
$$K = {{\left( {{K_1} + {K_2}} \right)\left( {{K_3} + {K_4}} \right)} \over {2\left( {{K_1} + {K_2} + {K_3} + {K_4}} \right)}}$$
C
$$K = {{\left( {{K_1} + {K_2}} \right)\left( {{K_3} + {K_4}} \right)} \over {{K_1} + {K_2} + {K_3} + {K_4}}}$$
D
$$K = {{\left( {{K_1} + {K_4}} \right)\left( {{K_2} + {K_3}} \right)} \over {2\left( {{K_1} + {K_2} + {K_3} + {K_4}} \right)}}$$
2
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
A parallel plate capacitor is made of two square plates of side 'a', separated by a distance d (d < < a). The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is :

A
$${{K{ \in _0}{a^2}} \over {2d(K + 1)}}$$
B
$${{K{ \in _0}{a^2}} \over {d(K - 1)}}\ln K$$
C
$${{K{ \in _0}{a^2}} \over d}\ln K$$
D
$${1 \over 2}{{K{ \in _0}{a^2}} \over d}$$
3
JEE Main 2018 (Online) 16th April Morning Slot
+4
-1
In the following circuit, the switch S is closed at t = 0. The charge on the capacitor C1 as a function of time will be given by $$\left( {{C_{eq}} = {{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)$$

A
$${C_1}E\left[ {1 - \exp \left( { - tR/{C_1}} \right)} \right]$$
B
$${C_2}E\left[ {1 - \exp \left( { - t/R{C_2}} \right)} \right]$$
C
$${C_{eq}}E\left[ {1 - \exp \left( { - t/R{C_{eq}}} \right)} \right]$$
D
$${C_{eq}}E\,\,\exp \left( { - t/R{C_{eq}}} \right)$$
4
JEE Main 2018 (Offline)
+4
-1
A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be :
A
0.9 n C
B
1.2 n C
C
0.3 n C
D
2.4 n C
EXAM MAP
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