Home
Class 12
PHYSICS
(a) Show that the normal component of el...

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
`(E_(2)-E_(1)).n = sigma/epsilon_(0)`
where `hatn` is a unit vector normal to the surface at a point and `sigma` is the surface charge density at that point. (The direction of `hatn` is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is `sigma hatn //epsilon_(0)`.
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

Answer

Step by step text solution for (a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by (E_(2)-E_(1)).n = sigma/epsilon_(0) where hatn is a unit vector normal to the surface at a point and sigma is the surface charge density at that point. (The direction of hatn is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is sigma hatn //epsilon_(0). (b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.] by PHYSICS experts to help you in doubts & scoring excellent marks in Class 12 exams.

Doubtnut Promotions Banner Mobile Dark
|

Similar Questions

Explore conceptually related problems

Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by (E_2-E_1). overset(^^)(n)=(sigma)/(epsi_0) Where overset(^^)(n) is a nuit vector normal to the surface at a point and sigma is the surface charge density at that point . (The direction of overset(^^)(n) is from side 1 to side 2.) Hence show that just out side a conductor , the electric field is sigma overset(^^)(n)//epsi_0 . (b) Show that the tangential component of electrostatic field is continuous from one side of a charge surface of another. use the fact that work done by electrostatic field on a closed loop is zero. )

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (sigma//epsi epsi_(0))hat(n) , where hat(n) is the unit vector in the outward normal direction and sigma is the surface charge density near the hole.

Knowledge Check

  • The magnitude of the electric field on the surface of a sphere of radius r having a uniform surface charge density sigma is

    A
    `sigma//epsi_(0)`
    B
    `sigma//2epsi_(0)`
    C
    `sigma//epsi_(0)r`
    D
    `sigma//2epsi_(0)r`
  • The magnitude of the electric field on the surface of a sphere of radius r having a uniform surface charge density sigma is

    A
    `sigma//epsilon_0`
    B
    `sigma//2epsilon_0`
    C
    `sigma//epsilon_0r`
    D
    `sigma//2epsilon_0r`
  • The magnitude of the electric field on the surface of a sphere of a sphere of radius r having a unifrom surface charge density sigma is

    A
    `sigma//epsilon_(0)`
    B
    `sigma//2epsilon_(0)`
    C
    `sigma//2epsilon_(0)r`
    D
    `sigma//2epsilon_(0)r`
  • Similar Questions

    Explore conceptually related problems

    We know that electric field is discontinuous across the surface of a charged conductor . Is electric potential also discontinuous there ?

    The surface charge density of a thin charged disc of radius R is sigma . The value of the electric field at the centre of the disc (sigma)/(2 epsilon_0) . With respect to the field at the centre, the electric field along the axis at a distance R from the centre of the disc is

    The surface charge density of a thin charged disc of radius R is sigma . The value of the electric field at the centre of the disc (sigma)/(2 in_(0)) . With respect to the field at the centre, the electric field along the axis at a distance R from the centre of the disc is

    (A ): Sharper is the curvature of spot on a charged body lesser will be the surface density of charge at that point. (R ): Electric field is zero inside a charged conductor.

    A long thin flat sheet has a uniform surface charge density sigma . The magnitude of the electric field at a distance .r. from it is given by