Hi. I am having a bit of trouble. If someone could show me thesteps on how to fi

Question

Hi. I am having a bit of trouble. If someone could show me thesteps on how to figure out this problem I would greatly appreciateit. Thanks.


A point charge of -6 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 +x axis)

(b) Calculate the magnitude and direction of the force on anelectron at x = -1 m, y = 0.
3 N
4°(counterclockwise from +x axis (a) Find the magnitude and direction of theelectric field at x = -1 m, y = 0.
1 N/C
2°(counterclockwise from +x axis)

(b) Calculate the magnitude and direction of the force on anelectron at x = -1 m, y = 0.
3 N
4°(counterclockwise from +x axis

Explanation / Answer

(a)the electric field due to -6 C charge E1 = k * (q1/r1^2) k = (1/4o) = 9 * 10^9 Nm^2/C^2,q1 = -6* 10^-6 C and r1^2 = [(1 - (-1))^2 + (3 - 0)^2] = 13 the electric field due to 12 C charge E2 = k * (q2/r2^2) q2 = 12 * 10^-6 C and r2^2 = [(2 - (-1))^2 + (-2 - 0)^2] =13 the net electric field at x = -1 m, y =0 E = (Ex + Ey)^(1/2) Ex = E1x + E2x and Ey = E1y + E2y E1x = E1 * cos1 and E2x = E2 * cos2 similarly,E1y = E1 * sin1 and E2y = E2 *sin2 Using trinometry tan1 = [(3 - 0)/(1 - (-1))] or 1 = 56.3o and tan2 = [(2 - (-1))/(-2 - 0)] or 2 = -56.3o the direction of the electric field at x = -1 m,y = 0 tan = (Ey/Ex) or = tan-1(Ey/Ex) (counterclockwise from+x axis)
(b)the force on electron at x = -1 m, y = 0 F1 = k * (q1 * e/r1^2) e = -1.6 * 10^-19 C and F2 = k * (q2 * e/r2^2) the net electric force on electron at x =-1 m, y = 0 F = (Fx + Fy)^(1/2) Fx = F1x + F2x and Fy = F1y + F2y F1x = F1 * cos1 and F2x = F2 * cos2 similarly,F1y = F1 * sin1 and F2y = F2 *sin2 the direction of the electric force at x = -1 m,y = 0 tan = (Fy/Fx) or = tan-1(Fy/Fx) (counterclockwisefrom +x axis)
the net electric force on electron at x =-1 m, y = 0 F = (Fx + Fy)^(1/2) Fx = F1x + F2x and Fy = F1y + F2y F1x = F1 * cos1 and F2x = F2 * cos2 similarly,F1y = F1 * sin1 and F2y = F2 *sin2 the direction of the electric force at x = -1 m,y = 0 tan = (Fy/Fx) or = tan-1(Fy/Fx) (counterclockwisefrom +x axis)
the direction of the electric force at x = -1 m,y = 0 tan = (Fy/Fx) or = tan-1(Fy/Fx) (counterclockwisefrom +x axis)

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