Strength of electric field on the axis of disc at a distance 3r/4 from centre is
Answers
Explanation:
Consider the disc of radius "a" with uniform charge density σ; E be the electric field at a point along the axis of the disc at a distance x from its center.
We can assume the charge distribution as a collection of concentric rings of charge. Consider one such ring of radius r and charge dq.
Area element dA=(2πr)dr= area of ring
dq=σdA=(2πσr)dr= charge of that ring
Due to symmetry, there is no vertical component of electric field at point P. There is only the horizontal component so, using the result from a ring of charge;
dE
x
=
(x
2
+r
2
)
3/2
k(dq)x
=
(x
2
+r
2
)
3/2
k(2πσr)drx
E
x
=∫
0
a
dE
x
=∫
0
a
(x
2
+r
2
)
3/2
(2πσrkdr)x
put x
2
+r
2
=t
2
Differentiating both sides,
2rdr=2tdt
At r=0;t=x
At r=a;t=
x
2
+a
2
So, E
x
=
2ε
0
σ
∫
x
x
2
+a
2
t
3
t
dt
=
2ε
0
σ
[
t
−1
]
x
x
2
+a
2
=
2ε
0
σ
[
x
1
−
x
2
+a
2
1
]
=
2ε
0
σ
⎣
⎢
⎢
⎢
⎡
1−
x
2
a
2
+1
1
⎦
⎥
⎥
⎥
⎤
is very small compared to 1.
So, E
x
=
2ε
0
σ
.