1
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
2
JEE Main 2018 (Online) 15th April Evening Slot
+4
-1
A parallel plate capacitor with area 200 cm2 and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is $$25 \times {10^{ - 6}}N,$$ the value of V is approximately : $$\left( {{ \in _o} = 8.85 \times {{10}^{ - 12}}{{{C^2}} \over {N.{m^2}}}} \right)$$
A
250 V
B
100 V
C
300 V
D
150 V
3
JEE Main 2018 (Online) 15th April Evening Slot
+4
-1
A capacitor C1 = 1.0 $$\mu$$F is charged up to a voltage V = 60 V by connecting it to battery B through switch (1). Now C1 is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $${C_2} = 3.0\mu F$$ and C3 = 6.0 $$\mu$$F through switch (2), as shown in the figure. The sum of final charges on C2 and C3 is :
A
40 $$\mu$$C
B
36 $$\mu$$C
C
20 $$\mu$$C
D
54 $$\mu$$C
4
JEE Main 2018 (Online) 15th April Morning Slot
+4
-1
The equivalent capacitance between $$A$$ and $$B$$ in the circuit given below, is :

A
$$2.4\,\mu F$$
B
$$4.9\,\mu F$$
C
$$3.6\,\mu F$$
D
$$5.4\,\mu F$$
EXAM MAP
Medical
NEET