Electric field on the axis of circular uniformly charged ring
Consider a circular ring of radius 'R' having charge 'Q' which is uniformly distributed. Let the linear charge density of the ring is λ.
Then electric field will be ----
As
we know, λ = Q/l
λ = Q/2πR,
λ= dQ/dl, so dQ= λdl ..........(1)
Q= λ(2πR) ..............(2)
cosφ= x/r
cosφ = x/√R2+x2
λ = Q/2πR,
λ= dQ/dl, so dQ= λdl ..........(1)
Q= λ(2πR) ..............(2)
cosφ= x/r
cosφ = x/√R2+x2
As we know,
dE
= dQ/4πЄ₀r2
dE
= λdl/4πЄ₀r2
dE=
λdl/4πЄ₀ (√R2+x2
)2
Net electric field at point p will be----
ഽdE = ഽ02πRdE cosφ
ഽdE = λcosφ/4πЄ₀ (R2+x2 )ഽ02πRdl
E= λx/4πЄ₀ (√R2+x2
)(R2+x2
)[l]02πR {since, cosφ = x/√R2+x2 }
E= λx2πR /4πЄ₀ (R2+x2 )3/2
E=QX/4πЄ₀ (R2+x2 )3/2 {since, Q=λ2πR}
E=QX/4πЄ₀ (R2+x2 )3/2 {since, Q=λ2πR}
so net electric field at point p will be------------
E = QX /4πЄ₀ (R2+x2 )3/2
Important conclusion : a) Electric field due to ring exist only in x-direction.
b)Net electric field due to ring in y-direction will be zero.
Ey =0
c) If point p is very far away from centre, i.e., x>>R,
x2
>>>>R2
so, R2
can be neglected, then the electric field will be-
E=Q/4πЄ₀x2
it means the ring will behave as a point charge.
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