1
JEE Main 2019 (Online) 11th January Morning Slot
+4
-1
In the figure shown below, the charge on the left plate of the 10$$\mu$$F capacitor is –30$$\mu$$C. The charge on the right plate of the 6 $$\mu$$F capacitor is :

A
+ 12 $$\mu$$C
B
+ 18 $$\mu$$C
C
$$-$$ 18 $$\mu$$C
D
$$-$$ 12 $$\mu$$C
2
JEE Main 2019 (Online) 10th January Evening Slot
+4
-1
A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is :
A
508 pJ
B
692 pJ
C
560 pJ
D
600 pJ
3
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
A parallel plate capacitor is of area 6 cm2 and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants K1 = 10, K2 = 12 and K3 = 14. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be -

A
12
B
36
C
14
D
4
4
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
$$K = {{\left( {{K_1} + {K_4}} \right)\left( {{K_2} + {K_3}} \right)} \over {2\left( {{K_1} + {K_2} + {K_3} + {K_4}} \right)}}$$
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