Home
Class 12
PHYSICS
A slab of a material of dielectric const...

A slab of a material of dielectric constaant 2 is placed between two identical parallel plates. One of the plates carries total charge density +2mC/`m^(2)` and the other plate carries total charge density -4mC/`m^(2)` the magnitude of electric field inside the dielectric in (in `10^(11)N//C`) Take `epsi_(0)=(80)/(9)xx10^(-12)Nm^(2)//C^(2)`)

A

2.25

B

3.38

C

4.5

D

6.75

Text Solution

Verified by Experts

The correct Answer is:
B
The inner surfaces of the two plates will carry charge density `+3mC//m^(2) and -3mC//m^(2)` respectively. Therefore, field between the plates,
`E=(sigma)/(epsi_(0))=(3)/((80)/(9)xx10^(-12))=3.38xx10^(11)N//C`
Promotional Banner

Similar Questions

Explore conceptually related problems

The surface charge density of a conductor is 12xx10^(-12) C//m^(2) .If the conductor is surrounded by a medium of dielectric constant 3.14, the magnitude of electric field just outside the conductor is ((1)/(4 piepsi_(0)) = 9xx10^(9) Nm^(2)//C^(2))

A cyclinder of radius 5 mm and surface charge density of 2 mu C//m^(2) is surrounded by a medium of dielectric constant 6.28. The magnitude of electric field at a point 2 m away from the axis of the cyclinder is ((1)/(4)piepsi_(0) = 9xx10^(9) Nm^(2)//C^(2))

The energy density in an electric field of intensity 200 "volt"//m , if K=4 and epsi_(0)=8.85 xx 10^(-12) C^(2)//Nm^(2) is

One plate of a capacitor of capacitance 2muF has total charge +10muC and its other plate has total charge +40mUC . The potential difference between the plates is (in Volt).

A parallel-plate air capacitor of plate separation 2mm and capacitance 1muF is charged to 100V. A dielectric slab of relative permittivity 50 is now inserted so as to fill the space between the plates. (i)Find the polarisation charge on one of the boundaries of the dielectric slab. (ii) Find the magnitude of the polarisation of the dielectric slab. (epsilon_(0) = 8.85 xx 10^(-12) C^(2)//N.m^(2))

A dielectric slab of thickness 1.0 cm and dielectric constant 5 is placed between the plates of a parallel plate capacitor of plate area 0.01 m^(2) and separation 2.0 cm. Calculate the change in capacity on introduction of dielectric. What would be on the change, if the dielectric slab were conducting?

The electric field due to a charge at a distance of 3 m from it is 500 NC^(-1) . The magnitude of the charge is [(1)/(4 pi epsi_(0)) = 9 xx 10^(9) N - m^(2)//C^(2)]

The electric potential difference between two parallel plates is 2000 V . If the plates are separated by 2 mm, then what is the magnitude of electrostatic force on a charge of 4 xx 10^(-6)C located midway between the plates ?

A dielectric of thickness 5cm and dielectric constant 10 is introduced between the plates of a parallel plate capacitor having plate area 500 sq. cm and separation between the plates 10cm. The capacitance of the capacitor with dielectric slab is (epsi_(0) = 8.8 xx 10^(-12)C^(2)//N-m^(2))