Question

derivation of electric field intensity at any point on the axis of a uniformly charged ring​

Answer 1

Answer:

This is the answer

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Answer 2

Answer:

expression of electric field on the axis of the ring is given by,

E = Kqx/(R² + x²)^(3/2) see the figure, here a ring is placed and a point located in its axis ( x unit away from the centre of the ring.) now cut an element of ring which makes an angle θ with line of axis