# Three Point Charges, find the electric force exerted

by m00nbeam360
Tags: charges, electric, exerted, force, point
 P: 20 1. The problem statement, all variables and given/known data Three charges, each of magnitude 2 nC, are at separate corners of a square of edge length 1 cm. The two charges at opposite corners are positive, and the other charge is negative. Find the force exerted by these charges on a fourth charge q = +3 nC at the remaining (upper right) corner. (Assume the +x axis is directed to the right and the +y axis is directed upward.) 2. Relevant equations k = (8.988*10^9 N*m^2/C^2) 3. The attempt at a solution Drew a diagram of the problem such that F4 = F1,4 + F2,4 + F 3,4. F1,4 = (k*q1*q4)/(r(1,4)^2)j F1,4 = (8.988*10^9 N*m^2/C^2)(2.00*10^-9)*((2.00*10^-9)/(0.01m)^2) = 3.595 * 10^-4 j F2,4 = same F3,4 = (k*q1*q2)/(r(2,4)^2)*r(3,4) r(3,4) = r(3,1) + r(1,4) = (0.01m)i + (0.01m)j r(3,4) = (0.01m)i+(0.01m)j/(sqrt(0.01m^2+0.01m^2)) = 0.14i + 0.14j F3,4 = (8.988*10^9) (-2.00*10^-9)(2.00*10^-9)/(0.01sqrt(2)^2)*(0.14i+0.14j) = -2.52 * 10 ^ -5 (3.595 * 10^-4) i - (-2.52*10^-5) i = 3.34*10^-4i <--- Wrong :( (3.595 * 10^-4) j - (-2.52*10^-5) j = 3.34*10^-4j <--- Wrong :( Any suggestions?? I think that somewhere in computing F3,4 I put in an incorrect number. Thanks!
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P: 3,686
 Quote by m00nbeam360 r(3,4) = (0.01m)i+(0.01m)j/(sqrt(0.01m^2+0.01m^2)) = 0.14i + 0.14j
Check this.
 P: 20 Whoa, got .707 now, thanks. So then is everything else correct?
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P: 3,686

## Three Point Charges, find the electric force exerted

 Quote by m00nbeam360 F1,4 = (k*q1*q4)/(r(1,4)^2)j F1,4 = (8.988*10^9 N*m^2/C^2)(2.00*10^-9)*((2.00*10^-9)/(0.01m)^2) = 3.595 * 10^-4 j
Did you use the correct value for q4 here?

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