Home
Class 12
PHYSICS
The sketech below show cross secion so e...

The sketech below show cross secion so equipotential surfaces between two charged.Conductors that are shwon in solid black.Some points on the wquipotenital surfaces .near the conductors are marked as `A,B,C…. .` The arrangements lies in air (Take `epsilon_(0)=8.85xx10^(-12)C^(2)//Nm^(2)]`

Surfaces charge density of the plate is equal to

A

`8.85xx10^(-10)C//m^(2)`

B

`-8.85xx10^(-10)C//m^(2)`

C

`17.7xx10^(-10)C//m^(-2)`

D

`-17.7xx10^(-10)C//m(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`E=(40-10)/(0.3)=100V//m`
(near the plate the electric field has to be uniform \it is alomost due to the particle
For conducting plate
`E=(sigma)/(epsilon_(0))sigma=epsilon_(0)E`
therefore `s=8.85xx10^(-12)xx100`
`=8.85xx10^(-10)C//m^(2)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTROSTATICS AND GAUSS LAW

    NARAYNA|Exercise Passage II|3 Videos

  • ELECTROSTATICS AND GAUSS LAW

    NARAYNA|Exercise Passage III|3 Videos

  • ELECTROSTATICS AND GAUSS LAW

    NARAYNA|Exercise Level V|93 Videos

  • ELECTROSTATIC POTENTIAL AND CAPACITANCE

    NARAYNA|Exercise ADDITIONAL PROBLEMS|14 Videos

  • EXPERIMENTAL PHYSICS

    NARAYNA|Exercise Comprehension type|6 Videos

Similar Questions

Explore conceptually related problems

A point charge of 4.427 mu C is at the centre of a cube Gaussian surface 9.0 cm on the edge. Find the net electric flux through the surface. ( epsilon_0=8.854xx10^(-12)C^2 N^(-1) m^(-2) )

The surface density of charge on the surface of a charged conductor in the air is 26.5 muCm^(-2) . The the outward force per unit area of the charged conductor is (epsilon_0=8.85xx10^-12C^2N^-1m^-2)

Knowledge Check

  • The sketech below show cross secion so equipotential surfaces between two charged.Conductors that are shwon in solid black.Some points on the wquipotenital surfaces .near the conductors are marked as A,B,C…. . The arrangements lies in air (Take epsilon_(0)=8.85xx10^(-12)C^(2)//Nm^(2)] A positive charge is placed at B ,When it is released:

    A
    no force will be exerted on it .
    B
    it will move towards `A`
    C
    it will move towards `C`
    D
    it will move towards `E`

  • The sketch below shows cross-sections of equipotential surface between two charged conductors that are shown in solid black. Some points one the euquipotential surface near the conductors are marked as A, B, C ………. The arrangement lies in air. ["Take " in_(0)=8.85xx10^(-12) C^(2)//N m^(2)] How much work is required to slowly move a -1 mu C charge from E to D?

    A
    `2xx10^(-5) J`
    B
    `-2xx10^(-5) J`
    C
    `4xx10^(-5) J`
    D
    `-4xx10^(-5) J`

  • The surface density of charge on the surface of a charged conductor in air is 0.885 muC//m^(2) . If epsi_(0) =8.85xx10^(-12) C^(2)//N-m^(2) , then the outward force per unit area of the charged conductor is

    A
    `5xx10^(-2) N//m^(2)`
    B
    `4.425xx10^(-2) N//m^(2)`
    C
    `8.85xx10^(-2) N//m^(2)`
    D
    `5xx10^(-3) N//m^(2)`

    • Similar Questions

    Explore conceptually related problems

    The displacement current density across a parallel plate air capacitor which is being charged at the rate of 10^(12) V//m//s is (epsilon_0 = 8.85 xx 10^(-12) C^2 N^(-1)m^(-2)) :

    A metal plate of area 2m^(2) is charged with 12xx10^(-6)C . The surface surface density of charge is

    A metal plate of surface area 2m^(2) is charged with sqrt(8.85muC) . What is the mechanical force acting on the plate if it is kept in air ? [ epsi_(0)=8.85xx10^(-12)C^(2) ]

    Each plate of a parallel plate condenser has surface area of 5 cm^(2) . The distance between the plates is 2 mm. The dielectric constant of the medium between the plates is 5. Then the capacity of the condenser is (epsi_(0) = 8.85 xx 10^(-12) C^(2)//Nm^(2))

    The surface density of charge on a conductor situated in air, is 2xx10^(-4)C//m^(2) . The electric intensity at a point very near to its surface is

    Home
    Profile