Question

Find the expression for electric field intensity at a point on the axis of a uniformly charged ring.

Answer 1

Have a look on the attachment


Hope it helps..

Answer 2

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.

so, dEx = Kdq/r² cosθ

here, r = √(R² + x²) and cosθ = x/r = x/√(R² + x²)

now, dEx = Kdq/(R² + x²) × x/√(x² + R²)

= Kdqx/(R² + x²)^(3/2)

now, E = ∫dEx = ∫Kdqx/(R² + x²)^(3/2)

= Kqx/(R² + x²)^(3/2)

hence, expression of electric field on the axis of the ring is given by, E = Kqx/(R² + x²)^(3/2)

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