A circular ring of radius r is uniformly charged with a linear charge density λ . Determine the electric potential at a point on the axis located at a distance x from the centre of the ring. Subsequently, use this

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Correct answer: A. rλ 2 _0 (r^2 + x^2)^ 1/2 , rλ x 2 _0 (r^2 + x^2)^ 3/2

Consider an elemental charge dq on the ring of radius r . The linear charge density is λ , so dq = λ dl . The total charge on the ring is Q = ∫ λ dl = λ (2π r) . The electric potential V at a point on the axis at a distance x from the centre is given by integrating the potential due to each elemental charge: V = ∫ (1)/(4π _0) dq r^2 + x^2 Since the distance r^2 + x^2 is constant for all points on the ring: V = 1 4π _0 r^2 + x^2 ∫ dq = Q 4π _0 r^2 + x^2 Substituting Q = 2π r λ : V = 2π r λ 4π _0 r^2 + x^2 = rλ 2 _0 (r^2 + x^2)^ 1/2 The electric field E along the axis is the negative gradient of the potential: E = -(dV)/(dx) = -(d)/(dx) ( (rλ)/(2 _0) ( r^2 + x²)^ -1/2 ) E = -(rλ)/(2 _0) ( -(1)/(2) ( r^2 + x^2)^ -3/2 × 2x ) E = rλ x 2 _0 (r^2 + x^2)^ 3/2

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