If electric field around a surface is given by `|vec(E)|=(Q_(in))/(epsilon_0|A|)` where 'A' is the normal area of surface and `Q_(in)` is the charge enclosed by the surface. This relation of gauss's law is valid when

lf the electric field around a surface is given by |vecE|=(Q)/(E_(0)|vecA|) where vecA is the normal area of surface and Q_("in") is the charge enclosed by the surface. This relation of Gauss's law is valid when

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) hatn= (sigma)/(epsilon_0) where n is a unit vector normal to the surface at a point and 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 . Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

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) hatn= (sigma)/(epsilon_0) where hatn is a unit vector normal to the surface at a point and is the surface charge density at that point. (The direction of hat n is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is sigma hatn // epsilon_0

Show that the electric field at the surface of a charged conductor is given bu vecE = sigma/epsilon_0 hatn , where sigma is the surface charge density and hatn is a unit vector normal to the surface in the outward direction.

(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.]

(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.]

Which of the following law gives a relation between the electric flux through any closed surface and the charge enclosed by the surface ?

In fiinding the electric field using Gauses law the formula |vec(E)|=(q_("enc"))/(in_(0)|A|) is applicable. In the Formula. in_(0) is permittivity of free space, A is the area of Gaussian surface and q_("enc") is charge enclosed by the Gaussion surface. This equation can be used in which of the following situation?

Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by (oversetrarr(E_2)-oversetrarr(E_1))overset^n=sigma//epsilon_0 where overset^n isa unit vecotr normal to the surface at a point and ct is the surface charge density at that point. (The direction of overset^n is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is sigmaoverset^n//epsilon_0 .