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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. ) |
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Answer» Solution :(a) Normal component of electric field INTENSITY due to a thin infinite plane sheet of charge on left side. `vec(E_1)=-(SIGMA)/(2epsi_0)overset(^^)n` and on right side `2=vec(E_2)=(sigma)/(2epsi_0)overset(^^)n` Discontinuity in the normal component from ONE side to the other is `vec(E_1)-vec(E_2)=(simga)/(2epsi_0)overset(^^)n+(sigma)/(2epsi_0)overset(^^)n=(sigma)/(epsi_0)overset(^^)n` or `(vec(E_2)-vec(E_1))overset(^^)n=(sigma)/(epsi_0)overset(^^)n.overset(^^)n=(sigma)/(epsi_0)` inside a closed conductor `vec(E)_1=0 E=vec(E)_2=(sigma)/(epsi_0)overset(^^)n` (b) To show that the tangential component of electrostatic field is CONTINUOUS from one side of a charged surface to another, we use the fact thatworkdone by electrostatic field on a closed loop is ZERO. |
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