Show that the electric field at a point on the surface of a charged conductor or just outside it is perpendicular to the surface?
Answers
Answer:
Let a charge Q be given to a conductor, this charge under electrostatic equilibrium will redistribute and the electric field inside the conductor is zero (i.e. E
in
=0).
Let us consider a point P at which electric field strength is to be calculated, just outside the surface of the conductor. Let the surface charge density on the surface of the conductor in the neighbourhood of P be σ coulomb/ metre
2
. Now consider a small cylindrical box CD having one base C passing through P; the other base D lying inside the conductor and the curved surface being perpendicular to the surface of the conductor.
Let the area of each flat base be a. As the surface of the conductor is equipotential surface, the electric field strength E at P, just outside the surface of the conductor is perpendicular to the surface of the conductor in the neighbourhood of P.
The flux of electric field through the curved surface of the box is zero, since there is no component of electric field E normal to curved surface. Also the flux of electric field through the base D is zero, as electric field strength inside the conductor is zero. Therefore the resultant flux of electric field through the entire surface of the box is same as the flux through the face C. This may be analytically seen as:
If S
1
and S
2
are flat surfaces at C and D and S
3
is curved surface, then
Total electric flux
∮
S
E
⋅
dS
=∮
S
1
E
⋅
dS
1
+∮
S
2
E
⋅
dS
2
+∮
S
3
E
⋅
dS
3
=∮
S
1
E dS
1
cos0+∮
S
2
0
⋅
dS
2
+∮
S
3
E dS
3
cos90
∘
∮
S
E dS
1
=She
As the charge enclosed by the cylinder is (σa) coulomb, we have, using Gauss's theorem, =
ϵ
0
1
×charge enclosed
Ea=
ϵ
0
1
(σa) or E=
ϵ
0
p
...(i)
Thus the electric field strength at any point close to the surface of a charged conductor of any shape is equal to 1/ϵ
0
times the surface charge density σ. This is known as Coulomb's law. The electric field strength is directed radially away from the conductor if σ is positive and towards the conductor if σ is negative.
If
n
^
is unit vector normal to surface in outward direction, then
E
=
ϵ
0
p
n
^
Obviously electric field strength near a plane conductor is twice of the electric field strength near a non-conducting thin sheet of charge.
solution