Answer to Question #265863 in Electricity and Magnetism for zila

Question

Three charges, +5Q, -5Q, and +3Q are located on the y-axis at y = +4a, y = 0, and y=-4a , respectively. The point P is on the x-axis at x = 3a. (a) How much energy did it take to assemble these charges? (b) What are the x, y, and 2 components of the electric field Ę at P? (c) What is the electric potential V at point P, taking V = 0 at infinity? (d) A fourth charge of +Q is brought to P from infinity. What are the x, y, and z components of the force F that is exerted on it by the other three charges? (e) How much work was done (by the external agent) in moving the fourth charge +Q from infinity to P?

Answer 1

(a)

$U_{tot} = \sum^N_{i=1,j=1} (\frac{kq_iq)j}{r_{ij}}) \\ U = \frac{5(5φ)(-5φ)}{4a} + \frac{k(5φ)(3φ)}{8a} + \frac{k(-5φ)(3φ)}{4a} \\ U = -\frac{25kφ^2}{4a} + \frac{15kφ^2}{8a} -\frac{15kφ^2}{4a} \\ U = -\frac{65kφ^2}{8a} \;J$

(b) Electric field components: $E_x, E_y, E_z$ at P

$E_x = (E_{+5φ}cos53 + E_{+3φ} cos 53 -E_{-5φ}) \hat{i} \\ E_x = (\frac{k(5φ)}{25a^2}(\frac{3}{5}) + \frac{k(3φ)}{25a^2}(\frac{3}{5}) -\frac{k(5φ)}{9a^2} ) (+\hat{i}) \\ E_x = (\frac{24kφ}{325a^2} -\frac{5kφ}{9a^2}) \hat{i} \\ E_x = \frac{-2909 kφ}{5625a^2} \hat{i} \\ E_y = (E_{+5φ} sin 53 -E_{+3φ} sin53) \hat{j} \\ E_y = (\frac{5(5φ)}{25a^2} (\frac{4}{5}) -\frac{k(3φ)}{25a^2} (\frac{4}{5})) \hat{j} \\ E_y = \frac{8kφ}{625a^2} \hat{j} \\ E_z = 0$

(c)

$V_p = \frac{k(5φ)}{5a} +\frac{k(-5φ)}{3a} + \frac{k(+3φ)}{5a} \\ V_p = \frac{kφ}{a} -\frac{5kφ}{3a} +\frac{3kφ}{5a} \\ V_p = \frac{-kφ}{15a}$

(d)

$F_x =+φ \vec{E_x} = \frac{-2909kφ^2 \hat{i}}{5625a^2} \\ F_y = +φ \vec{E_y} = \frac{+8kφ^2 \hat{j}}{625a^2} \\ F_z = 0$

(e)

$WD_{out}=ΔU = φΔV = φ(V_f -V_i) = φ(V_p -V_{\infty}) \\ WD_{ex} = \frac{-kφ^2}{15a}$