Obtain an expression for electric force
and electric pressure on the
surface of charged conductor
Answers
Answer:
Force on the Surface of Charged Conductor
The charge provided to a conductor is uniformly distributed on its surface. a repulsive force acts by the charge on the rest part from the the charge present at small element on the conductor, and this way, a force of repulsion acts at each small element on the conductor, and the total force on the surface of the conductor is the vector sum of force acting on all the small elements.
That is why the charged conductor experiences pressure outwards the charged conducting surface.
Let the surface charge density on the surface of the conductor be σ. We will consider points outside and inside the conductor, two identical poimts P
1
and P
2
respectively (see fig.2.20).
Since, the electric filed outside the charged conducting surface is
ε
0
σ
. Thus, electric field at point P
1
E
P1
=
ε
0
σ
.........(1)
Electric field inside the conductor is zero. Thus, electric field at P
2
E
P2
=0.........(2)
Now, we will divided the conductor into two parts:
(i) Element AB whose surface is ds and
(ii) The remaining part ACB
If the electric field intensity at near point due to element AB is
E
1
and
E
2
due to ACB part. Then, from the fig.2.20,
Ep
1
=E
1
+E
2
......(3)
(E
1
and E
2
are in same direction at P
1
)
and Ep
2
=E
1
−E
2
.......(4)
(E
1
and E
2
are in opposite direction at P
2
)
From eqns. (2) and (4)
E
1
−E
2
=0
i.e.,E
1
=E
2
........(5)
From eqns.(1),(3) and (5)
E
2
+E
2
=
ε
0
σ
or E
2
=
2ε
0
σ
............(6)
Thus,electric field intensity at element AB due to part ACB is
2πε
0
σ
. If total charge on element AB is dq, then force on element dF=E
2
dq=
2ε
0
σ
dq(∵dq=σds)
Thus, dF=
2ε
0
σ
2
ds=
2
1
ε
0
E
2
ds...........(7){∵E=
ε
0
σ
thus,σ=ε
0
E}
Force acting on whole of the surface,
F=∮
S
2ε
0
σ
2
ds=∫
S
2
ε
0
E
2
ds..(8)
F=
2
ε
0
E
2
∮
S
ds
Pressure on the unit area of surface
P=
ds
dF
=
2ε
0
σ
2
=
2
1
ε
0
E
2
.......(9)
This pressure is called electric pressure.
Explanation:
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