Question

Positive charge Q is distributed uniformly along x-axis from x=0

Positive charge Q is distributed uniformly along the x-axis from x = 0 to x = a. A positive point charge q is located on the positive x-axis at x = a + r, a distance r to the right of the end of Q (Fig. P21.89). Calculate the x- and y-components of the electric field produced by the charge distribution Q at points on the positive x-axis where x > a. Calculate the force (magnitude and direction) that the charge distribution Q exerts on q. Show that if r >> a, the magnitude of the force in part (b) is approximately Qq/4 pi epsilon_0 r^2. Explain why this result is obtained.

Answer 1

(a)Let us consider a small charge distribution dQ of length dx. The electric field due to the small element is, kdo kid dE So, the electric field along the horizontal direction is, kick Q+X S dE= 2-3 hotel ko 1 1 a X a+x Hence, the x component of electric field is Ex kg (1 1 a x a+x The electric field along vertical direction due to small charge element should be zero. Hence, the y component of electric field is E, = 0. (b)The magnitude of the force is, F =qE = QE kqQ/1 1 a x a+x For x=r, a 1 a a+r kQq1 1 F= 1 ar kQq1 Hence, the magnitude of the force is F=- The direction of force is along positive x axis. (c)for r>>, the force is, kol 1 1 F a+1 kol a+r-r r(a+r) a a kQq a a r(a+r) kola a r(r) 1 09 478, r Since a is very small. So, atr=r 1 99 Hence, the force is F 12 4718