Question

Question 2.16: (a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by Where is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ (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.]

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-88

Answer 1

normal component of electric field intensity in case of thin infinte plane sheet of charge , on left side (assume side - 1) $E_1=-\frac{\sigma}{2\epsilon_0}\bf{n}$
and on right side (side - 2),$E_2=\frac{\sigma}{2\epsilon_0}\bf{n}$
[ because we know, electric field produce in case of thin infinte plane sheet is $\frac{\sigma}{\epsilon_0}$ and also we know, sheet contains charge both sides on it ,then it divide equal parts of electric field for both side . that's why electric fields are directed left and right side of sheet of magnitude $\frac{\sigma}{2\epsilon_0}$ ]

discontinuity in the normal component from one side to the other is
$E_2-E_1=\frac{\sigma}{2\epsilon_0}\bf{n}+\frac{\sigma}{2\epsilon_0}\bf{n}=\frac{\sigma}{\epsilon_0}\bf{n}$
$(E_2-E_1).\bf{n}=\frac{\sigma}{\epsilon_0}\bf{n.n}=\frac{\sigma}{\epsilon_0}$

inside the conductor , $E_1$=0
so, E = $E_2$=$\frac{\sigma}{\epsilon_0}\bf{n}$

(b) we know, wordone by electrostatic field on a closed loop is zero. it is because potential at every point on charged surface remains constant.
means, if we have to she the tangential component of the electric field is continuous from one side of charhed surface to another , we use the fact that work done by electrostatic field on a closed loop is zero.