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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
Questions With Solutions

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Question : 16
Total: 37
Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by (
→
E2
−
→
E1
).
^
n
 =
σ
ε0
 where
^
n
 is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of
^
n
 is from side 1 to side 2). Hence show that just outside a conductor, the electric field is σ
^
n
ε0
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
Solution:  
Normal component of electric field intensity due to a thin infinite plane sheet of charge, on left side (side 1)
→
E1
 = -
σ
2ε0
^
n
and on right side (side 2),
→
E2
 =
σ
2ε0
^
n
Discontinuity in the normal component from one side to the other is
→
E2
−
→
E1
 =
σ
2ε0
^
n
+
σ
2ε0
^
n
 =
σ
ε0
^
n
 or (
→
E2
−
→
E1
) .
^
n
 =
σ
ε0
^
n
.
^
n
 =
σ
ε0
Inside a closed conductor,
→
E1
 = 0
∴ E =
→
E2
 =
σ
ε0
^
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 that work done by electrostatic field on a closed loop is zero.
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