In Figure ( a ), a particle of charge + Q produces an electric field of magnitud

Question

In Figure  (a), a particle of charge +Q produces an electric field of magnitude Epart at point P, at distance R from the particle. In Figure (b), that same amount of charge is spread uniformly along a circular arc that has radius R and subtends an angle ?. The charge on the arc produces an electric field of magnitude Earc at its center of curvature P. For what value of ? does Earc = 0.856Epart? (Hint: You will probably resort to a graphical solution.)

___________rad

In Figure (a), a particle of charge +Q produces an electric field of magnitude Epart at point P, at distance R from the particle. In Figure (b), that same amount of charge is spread uniformly along a circular arc that has radius R and subtends an angle ?. The charge on the arc produces an electric field of magnitude Earc at its center of curvature P. For what value of ? does Earc = 0.856Epart? (Hint: You will probably resort to a graphical solution.) ___________rad

Explanation / Answer

Epart = k Q/R^2

Earc = (k lamda/R) (2 sin(theta/2))

lamda = Q/(R theta)

==> Earc = (k Q/(R^2 theta)) (2 sin(theta)/2)

Earc = 0.856 Epart

(k Q/(R^2 theta)) (2 sin(theta)/2) = 0.856 k Q/R^2

(1/(theta)) (2 sin(theta/2)) = 0.856

sin(theta/2) = 0.428 theta

===> theta = 1.90 rad

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