A point charge of -3 µC is locatedat x = 2 m, y = -2 m. Asecond point charge of

Question

A point charge of -3 µC is locatedat x = 2 m, y = -2 m. Asecond point charge of 12 µC is located at x = 1 m,y = 3 m. (a) Find the magnitude and direction of theelectric field at x = -1 m, y = 0.
1 N/C
2° (counterclockwise from +xaxis)

(b) Calculate the magnitude and direction of the force on anelectron at x = -1 m, y = 0.
3 N
4° (counterclockwise from +xaxis) I KNOW HOW TO DO IT, I JUST GET A WRONG NUMBERWHEN I PUT IT IN THE COMPUTER, I WAS WONDERING IF YOU CAN GIVE ME ANUMBER FOR AN ANSEAR SO I SEE IF IT IS RIGHT, WILL RATE LIFESAVOREVEN IF WRONG (a) Find the magnitude and direction of theelectric field at x = -1 m, y = 0.
1 N/C
2° (counterclockwise from +xaxis)

(b) Calculate the magnitude and direction of the force on anelectron at x = -1 m, y = 0.
3 N
4° (counterclockwise from +xaxis) I KNOW HOW TO DO IT, I JUST GET A WRONG NUMBERWHEN I PUT IT IN THE COMPUTER, I WAS WONDERING IF YOU CAN GIVE ME ANUMBER FOR AN ANSEAR SO I SEE IF IT IS RIGHT, WILL RATE LIFESAVOREVEN IF WRONG

Explanation / Answer

(a)electric field due to first charge E1 = (1/4o) * (q/r^2) (1/4o) = 9 * 10^9 Nm^2/C^2,q = -3C = -3 * 10^-6 C and r = [(2 - (-1))^2 + (-2 - 0)^2]^(1/2) =13 electric field due to second charge E2 = (1/4o) * (q1/r1^2) q1 = 12 C = 12 * 10^-6 C and r1 = [(1 - (-1))^2 + (3 -0)^2]^(1/2) = 13 The net electric field is E = (Ex + Ey)^(1/2) Ex = E1x + E2x Ey = E1y + E2y E1x = E1 * cos E2x = E2 * cos1 E1y = E1 * sin E2y = E2 * sin1 cos = (-2/13) and sin =(13/13) = 1 cos1 = (3/13) and sin1 =(13/13) = 1 angle made by resultant electric field tan = (Ey/Ex) (b)force due to first charge F1 = (1/4o) * (q * e/r^2) e = 1.6 * 10^-19 C and r = 13 and F2 = (1/4o) * (q1 * e/r1^2) r1 = 13 The net electric force is F = (Fx + Fy)^(1/2) Fx = F1x + F2x Fy = F1y + F2y F1x = F1 * cos and F2x = F2 * cos1 F1y = F1 * sin and F2y = F2* sin1 angle made resultant force tan = (Fy/Fx) E2 = (1/4o) * (q1/r1^2) q1 = 12 C = 12 * 10^-6 C and r1 = [(1 - (-1))^2 + (3 -0)^2]^(1/2) = 13 The net electric field is E = (Ex + Ey)^(1/2) Ex = E1x + E2x Ey = E1y + E2y E1x = E1 * cos E2x = E2 * cos1 E1y = E1 * sin E2y = E2 * sin1 cos = (-2/13) and sin =(13/13) = 1 cos1 = (3/13) and sin1 =(13/13) = 1 angle made by resultant electric field tan = (Ey/Ex) (b)force due to first charge F1 = (1/4o) * (q * e/r^2) e = 1.6 * 10^-19 C and r = 13 and F2 = (1/4o) * (q1 * e/r1^2) r1 = 13 The net electric force is F = (Fx + Fy)^(1/2) Fx = F1x + F2x Fy = F1y + F2y F1x = F1 * cos and F2x = F2 * cos1 F1y = F1 * sin and F2y = F2* sin1 angle made resultant force tan = (Fy/Fx) The net electric force is F = (Fx + Fy)^(1/2) Fx = F1x + F2x Fy = F1y + F2y F1x = F1 * cos and F2x = F2 * cos1 F1y = F1 * sin and F2y = F2* sin1 angle made resultant force tan = (Fy/Fx)

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