Partially Filled Dielectrics
Trending Questions
Q. Two thin dielectric slabs of dielectric constants K1 and K2(K1<K2) are inserted between plates of a parallel plate capacitor, as shown in figure. The variation of electric field E between the plates with distance d as measured from plate P is correctly shown by
Q. A parallel-plate capacitor is connected across a battery. Now, keeping the battery connected, a dielectric slab is inserted between the plates. In this process,
- no work is done
- work is done by the battery and the stored energy increases
- work is done by the external agent and the stored energy decreases
- work is done by the battery as well as external agent but the stored energy does not change
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 capacitor is connected across a battery. Now, keeping the battery connected, a dielectric slab is inserted between the plates. In this process,
- no work is done
- work is done by the battery and the stored energy increases
- work is done by the external agent and the stored energy decreases
- work is done by the battery as well as external agent but the stored energy does not change
Q. For configuration of media of permittivity ε0, ε, ε0 between parallel plates each of area A as shown in the figure; the equivalent capacitance is
- ε0Ad
- εε0Ad
- εε0Ad(ε+ε0)
- εε0A(2ε+ε0)d
Q. A parallel-plate capacitor with no dielectric has a capacitance of 0.5 μF. The space between the plates is filled with equal amounts of two dielectric materials of dielectric constants 2 and 3 as shown in the figure. Find the capacitance of the system.
- 1.2 μF
- 1.8 μF
- 1.25 μF
- None of these
Q. A dielectric slab of area A and thickness d is inserted between the plates of a capacitor of area 2A with constant speed v as shown in the figure. Distance between the plates is d.
The capacitor is connected to a battery of e.m.f. E; the current in the circuit varies with time as
The capacitor is connected to a battery of e.m.f. E; the current in the circuit varies with time as
- none of these
Q. For configuration of media of permittivity ε0, ε, ε0 between parallel plates each of area A as shown in the figure; the equivalent capacitance is
- ε0Ad
- εε0Ad
- εε0Ad(ε+ε0)
- εε0A(2ε+ε0)d
Q.
Find the effective capacitance of the following system.
- ϵ0Ak1k2(k1+k2)d
- (k1+k2)ϵ0Ad
- ϵ0k1k2A(k1−k2)dln(k1k2)
- ϵ0(k1+k2)Adln(k1k2)
Q. One plate of a capacitor is fixed and the other is connected to a spring as shown in the figure. Area of both the plates is A. In steady state (equilibrium), separation between the plates is 0.8d (spring was unstretched and the distance between the plates was d when the capacitor was uncharged). The force constant of the spring is approximately
- 12532∈0AE2d3
- 2∈0AE2d3
- 6∈0E2Ad2
- ∈0AE32d3
Q. Find ceff
- kϵ0Ad
- ϵ0Ad(k+1k)
- ϵ0Ad(2k+1k+1)
- ϵ0A2d(3k+1k+1)
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.
Q. A parallel plate capacitor contains a mica sheet (thickness =10−3 m) and a sheet of fibre (thickness =0.5×10−3 m). The dielectric constant of mica is 8 and that of fibre is 2.5. Assuming that the fibre breaks down when subjected to an electric field of 6.4×106 V/m, find the maximum safe voltage that can be applied to the capacitor.
- 2.08 kV
- 2.08 V
- 9.6 kV
- 9.6 V