Partially Filled Dielectrics
Trending Questions
Q.
What causes damping?
Q. A capacitor cannot be used as a battery because
- It cannot store a large amount of charge
- It produces too much heat
- It gets discharged very rapidly
- It is very costly as compared to a battery
Q. Three capacitors, each of capacitance 4 μF are connected to a battery of voltage 10 V. When key K is closed, the charge which will flow through the battery is
- 40 μC
- 40/3 μC
- 80 μC
- 20 μC
Q. Capacitance of a parallel plate capacitor becomes 43 times its original value if a dielectric slab of thickness t=d2 is inserted between the plates [d is the separation between the plates]. The dielectric constant of the slab is
- 4
- 8
- 2
- 6
Q. A parallel plate air capacitor is made using two square plates each of side 0.2 m, spaced 1 cm apart. It is connected to a 50 V battery. What is the charge on each plate?
- 1.77×10−3μC
- 1.77×10−3C
- 3.54×10−3C
- 3.54×10−3μC
Q. Determine the current flowing (in A) through an element at time t=2 sec, if the charge flow is given by q=(2t2+3) C.
- 8
- 1
- 2
- 4
Q. 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. The capacitance of this capacitor is :
- 12Kϵ0a2d
- Kϵ0a2dlnK
- Kϵ0a2d(K−1)lnK
- Kϵ0a22d(K+1)
Q. In the circuit shown in Fig, capacitors A and B have identical geometry, but a material of dielectric constant 3 is present between the plates of B. The potential difference across A and B are respectively
- 2.5 V, 7.5 V
- 2 V, 8 V
- 7.5 V, 2.5 V
- 8 V, 2 V
Q. A parallel-plate capacitor has a dielectric slab in it. The slab just fills the space inside the capacitor. The capacitor is charged by a battery and then the battery is disconnected. Now, the slab is pulled out slowly at t=0. If at time t, the capacitance of the capacitor is C and potential difference between the plates of the capacitor is V, then which of the following graphs is/are correct.