1
JEE Main 2020 (Online) 2nd September Evening Slot
+4
-1
A 10 $$\mu$$F capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is :
A
20 $$\mu$$F
B
15 $$\mu$$F
C
10 $$\mu$$F
D
30 $$\mu$$F
2
JEE Main 2020 (Online) 8th January Evening Slot
+4
-1
A capacitor is made of two square plates each of side 'a' making a very small angle $$\alpha$$ between them, as shown in figure. The capacitance will be close to :
A
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 + {{\alpha a} \over {d}}} \right)$$
B
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {4d}}} \right)$$
C
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {2d}}} \right)$$
D
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{3\alpha a} \over {2d}}} \right)$$
3
JEE Main 2020 (Online) 8th January Morning Slot
+4
-1
Effective capacitance of parallel combination of two capacitors C1 and C2 is 10 μF. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor C2 is 4 times that of C1. If these capacitors are connected in series, their effective capacitance will be :
A
4.2 μF
B
8.4 μF
C
1.6 μF
D
3.2 μF
4
JEE Main 2020 (Online) 7th January Morning Slot
+4
-1
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
$${{A{ \in _0}K} \over d}\left( {1 + \alpha d} \right)$$
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