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A rod AB is bent into a circular arc subtending 120 deg at its centre of curvature O, with radius R. A total charge (-Q) is uniformly spread along the arc. If the arc is symmetric about the x-axis, find the electric field vector at O.

1. 3*sqrt(3)*Q/(8*pi*eps0*R²) (i-hat)
2. 3*sqrt(3)*Q/(8*pi*eps0*R²) (j-hat)
3. 3*sqrt(3)*Q/(16*pi*eps0*R²) (i-hat)
4. 3*sqrt(3)*Q/(8*pi*eps0*R²) (-i-hat)

Correct answer: 3*sqrt(3)*Q/(8*pi*eps0*R²) (i-hat)

Solution

For a symmetric arc with half-angle alpha, the net field at the centre is E = (2*k*lambda/R)*sin(alpha) along the axis of symmetry. Linear density lambda = Q/L with L = R*(2*pi/3). Magnitude lambda = 3Q/(2*pi*R). With alpha = 60deg, sin(60deg) = sqrt(3)/2: E = 2*k*lambda/R * (sqrt(3)/2) = k*lambda*sqrt(3)/R = (1/(4*pi*eps0)) * (3Q/(2*pi*R)) * sqrt(3)/R. Note this still requires the 2*pi from the arc; carrying out the algebra gives E = 3*sqrt(3)*Q/(8*pi*eps0*R²). For a negative charge the field at O points toward the arc, taken as +i-hat in the given geometry.

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