Four charges Aq,Bq,Cq, and Dq (q = 1.50 times 10^7C) sit in a plane at the corne

Question

Four charges Aq,Bq,Cq, and Dq (q = 1.50 times 10^7C) sit in a plane at the corners of a square whose sides have length d = 26.0 cm, as shown in the diagram below. A charge, Eq, is placed at the origin at the center of the square. DATA: A = 8, B = 4, C = 5, D = 2, E = 4. Consider the charge at the center of the square. What is the net force, in the x-direction, on this charge? Computer's answer now shown above. Tries 0/15 What is the net force, in the y-direction, on the center charge? Computer's answer now shown above.

Explanation / Answer

For this problem, calculate the force that each surrounding charge has on the charge in the center. Since the sides of the square have length d, the distance that each surrounding charge is from the center is

r = sqrt[(d/2)^2 + (d/2)^2] = sqrt[2(d/2)^2] = d/sqrt(2)

FD=(Bq*Eq*K)/(r^2) = (9*10^9*4*1.5*10^-7*4*1.5*10^-7)/(0.26^2/2) = 0.096 N
FA=(Cq*Eq*K)/(r^2) = (9*10^9*5*1.5*10^-7*4*1.5*10^-7)/(0.26^2/2) = 0.12 N
FC=(Aq*Eq*K)/(r^2) = (9*10^9*8*1.5*10^-7*4*1.5*10^-7)/(0.26^2/2) = 0.19 N
FB=(Dq*Eq*K)/(r^2) = (9*10^9*2*1.5*10^-7*4*1.5*10^-7)/(0.26^2/2) = 0.048 N

Now a) Fx= -FDcos45* - FAcos45* + FBcos45* + FCcos45* = cos45(-FD -FA + FB + FC) = 15.56 mN

b) Fy= -FCsin45* - FDsin45* + FAsin45* + FBsin45* = sin45(-FC -FD + FA + FB) = -0.118 N

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