This WebAssign question is not too complex, but I keep getting thewrong answer,

Question

This WebAssign question is not too complex, but I keep getting thewrong answer, please help.

Two 3.00 µC charged particles arelocated on the x axis. One is at x = 1.00 m, andthe other is at x = -1.00 m. (a) Determine the electric field on they axis at y = 0.500m.

N/Ci + N/Cj

(b) Calculate the electric force on a -3.00 µC charge placedon the y axis at y = 0.500 m.
N i+ Nj
I have been using the formula E=kQ/r2


(a) Determine the electric field on they axis at y = 0.500m.

N/Ci + N/Cj

(b) Calculate the electric force on a -3.00 µC charge placedon the y axis at y = 0.500 m.
N i+ Nj

Explanation / Answer

(a)the electric field due to the charge at x = 1.00 m at y =0.500 m is E1 = k * (q/r2) k = (1/4o) = 9 * 109Nm2/C2,q = 3.00 C = 3.00 *10-6 C and r2 = ((1.00)2 +(0.500)2) = 1.25 m2 the electric field due to charge at x = -1.00 m at y = 0.500 mis E2 = k * (q/r2) the electric field due to charge at x = -1.00 m at y = 0.500 mis E2 = k * (q/r2) r2 = ((-1.00)2 +(0.500)2) = 1.25 m2 the net electric field on the y axis at y = 0.500 m is E = (Ex2 +Ey2)1/2 Ex = E1x + E2x =E1 * cos1 + E2 *cos2 and Ey = E1y + E2y =E1 * sin1 + E2 *sin2 the angles 1 and 2 are tan1= (0.500/1.00) or 1= tan-1(0.500) =26.5o and tan2 = (0.500/-1.00) or 2= tan-1(-0.500) =-26.5o the electric field can also be written as E = iEx + jEy (b)the electric force on a -3.00 C charge placed on they axis at y = 0.500 m is F1= k *(q1q3/r2) q1 = 3.00 C = 3.00 * 10-6 C andq3= -3.00 C = -3.00 * 10-6 C similarly,the electric force between charge q3and the other charge is F2= k *(q2q3/r2) q2 = 3.00 C = 3.00 * 10-6 C the net electric force on the charge at y = 0.500 m is F = (Fx2 +Fy2)2 Fx= F1x + F2x = F1* cos1 + F2 *cos2 and Fy = F1y + F2y =F1 * sin1 + F2 *sin2 the angles 1 and 2 are26.5o and -26.5o respectively. the electric force can also be written as F = iFx + jFy the angles 1 and 2 are26.5o and -26.5o respectively. the electric force can also be written as F = iFx + jFy

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