Question

Derive electric field on the axis of a charged Ring.​

Answer 1

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)

Answer 2

Answer:

refer the attachment file mate!☺