A rod of length L carries a total charge Q distributed uniformly along its length. The rod is bent into the shape of a semicircle. Determine the magnitude of the electric field at the centre of curvature of the semicircle.

Step-by-step solution

Let the radius of the semicircle be R . Since the rod of length L is bent into a semicircle, the length of the arc is equal to L . R = L R = L The linear charge density of the rod is = Q L . Consider a small element of the semicircle subtending an angle d at the centre. The charge on this element is dq = R d . The electric field due to this element at the centre is dE = 1 4 ₀ dq R^2 = 1 4 ₀ R d R^2 = 4 ₀ R d . By symmetry, the components of the electric field perpendicular to the angle bisector cancel out. The net