1
JEE Main 2020 (Online) 7th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
JEE Main 2020 (Online) 7th January Morning Slot Physics - Capacitor Question 96 English A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1 + $$\alpha $$x) where 'x' is the distance measured from one of the plates. If (ad) << 1, the total capacitance of the system is best given by the expression :
A
$${{A{ \in _0}K} \over d}\left( {1 + {{\left( {{{\alpha d} \over 2}} \right)}^2}} \right)$$
B
$${{A{ \in _0}K} \over d}\left( {1 + {{\alpha d} \over 2}} \right)$$
C
$${{A{ \in _0}K} \over d}\left( {1 + {{{\alpha ^2}{d^2}} \over 2}} \right)$$
D
$${{A{ \in _0}K} \over d}\left( {1 + \alpha d} \right)$$
2
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In the given circuit, the charge on 4 $$\mu $$F capacitor will be : JEE Main 2019 (Online) 12th April Evening Slot Physics - Capacitor Question 97 English
A
5.4 $$\mu $$C
B
9.6 $$\mu $$C
C
13.4 $$\mu $$C
D
24 $$\mu $$C
3
JEE Main 2019 (Online) 12th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K1, K2 and K3. The first capacitor is filled as shown in fig.I, and the second one is filled as shown in fig II. If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be (E1 refers to capacitor (I) and E2 to capacitor (II)): JEE Main 2019 (Online) 12th April Morning Slot Physics - Capacitor Question 98 English
A
$${{{E_1}} \over {{E_2}}} = {{\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)} \over {{K_1}{K_2}{K_3}}}$$
B
$${{{E_1}} \over {{E_2}}} = {{{K_1}{K_2}{K_3}} \over {\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}$$
C
$${{{E_1}} \over {{E_2}}} = {{\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)} \over {9{K_1}{K_2}{K_3}}}$$
D
$${{{E_1}} \over {{E_2}}} = {{9{K_1}{K_2}{K_3}} \over {\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}$$
4
JEE Main 2019 (Online) 10th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E, as shown in figure. Its bob has mass m and charge q. The time period of the pendulum is given by : JEE Main 2019 (Online) 10th April Evening Slot Physics - Capacitor Question 99 English
A
$$2\pi \sqrt {{L \over {\sqrt {{g^2} - {{{q^2}{E^2}} \over {{m^2}}}} }}} $$
B
$$2\pi \sqrt {{L \over {\left( {g + {{qE} \over m}} \right)}}} $$
C
$$2\pi \sqrt {{L \over {\sqrt {{g^2} + {{{q^2}{E^2}} \over {{m^2}}}} }}} $$
D
$$2\pi \sqrt {{L \over {\left( {g - {{qE} \over m}} \right)}}} $$
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