A ring of radius a contains a charge q distributed uniformly over its length. Find the electric field at a point on the axis of the ring at a distance x from the centre.
Answers
Answer:
Consider a point P on the axis of uniformly charged ring at a distance x from its centre O. Point P is at distance r=
a
2
+x
2
from each element dl of ring. If q is total charge on ring, then, charge per metre length, λ=
2πa
q
.
The ring may be supposed to be formed of a large number of ring elements.
Consider an element of length dl situated at A.
The charge on element, dq=λdl
∴ The electric field at P due to this element
dE
1
=
4πϵ
0
1
r
2
dq
=
4πϵ
0
1
r
2
λdl
, along
PC
The electric field strength due to opposite symmetrical element of length dl at B is
dE
2
=
4πϵ
0
1
r
2
dq
=
4πϵ
0
1
r
2
λdl
, along
PD
If we resolve
dE
1
and
dE
2
along the axis and perpendicular to axis, we note that the components perpendicular to axis are oppositely directed and so get cancelled, while those along the axis are added up. Hence, due to symmetry of the ring, the electric field strength is directed along the axis.
The electric field strength due to charge element of length dl, situated at A, along the axis will be
dE=dE
1
cosθ=
4πϵ
0
1
r
2
λdl
cosθ
But, cosθ=
r
x
∴dE=
4πϵ
0
1
r
3
λdlx
=
4πϵ
0
1
r
3
λx
dl
The resultant electric field along the axis will be obtained by adding fields due to all elements of the ring, i.e.,
∴E=∫
4πϵ
0
1
r
3
λx
dl=
4πϵ
0
1
∫dl
But, ∫dl= whole length of ring =2πa and r=(a
2
+x
2
)
1/2
∴E=
4πϵ
0
1
(a
2
+x
2
)
3/2
λx
2πa
As, λ=
2πa
q
, we have E=
4πϵ
0
1
(a
2
+x
2
)
3/2
(
2πa
q
)x
2πaE=
4πϵ
0
1
(a
2
+x
2
)
3/2
qx
or, E=
4πϵ
0
1
(a
2
+x
2
)
3/2
qx
, along the axis
At large distances i.e., x>>a,E=
4πϵ
0
1
x
2
q
i.e., the electric field due to a point charge at a distance x.
For points on the axis at distances much larger than the radius of ring, the ring behaves like a point charge.
Explanation:
