find the expression of electric field intensity at a point on the axis of a uniformly charged ring
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We have a ring of radius ‘a’ and a charge ‘q’ distributed over it uniformly.
Let us consider a small element of the ring at the point A having charge dq.
The field at P due to this element is given as: dE=kdq/AP²=kdq/(a²+x²)
By symmetry the field at P due to the points on the ring will be along OP.
The perpendicular components will be cancelled out due to diametrically opposite points on the ring.
The component of dE along OP is,
dE cos Ѳ= kdq/(a²+x²) OP/AP= kdq/(a²+x²) x/(a²+x²)1/2
dE cos Ѳ= kxdq/(a²+x²)3/2
So the net field at P is,
E=∫dE cos Ѳ= ∫ kxdq/(a²+x²)3/2
E= kxq/(a²+x²)3/2
E= xq/4πε₀(a²+x²)3/2
Let us consider a small element of the ring at the point A having charge dq.
The field at P due to this element is given as: dE=kdq/AP²=kdq/(a²+x²)
By symmetry the field at P due to the points on the ring will be along OP.
The perpendicular components will be cancelled out due to diametrically opposite points on the ring.
The component of dE along OP is,
dE cos Ѳ= kdq/(a²+x²) OP/AP= kdq/(a²+x²) x/(a²+x²)1/2
dE cos Ѳ= kxdq/(a²+x²)3/2
So the net field at P is,
E=∫dE cos Ѳ= ∫ kxdq/(a²+x²)3/2
E= kxq/(a²+x²)3/2
E= xq/4πε₀(a²+x²)3/2
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