What Is the Correct Calculation for Net Force on Charge 3 in an Electric Field?

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SUMMARY

The correct calculation for the net force acting on charge 3, located at x = -1.080m with a charge of 45.5nC, involves applying Coulomb's Law. The forces from charge q1 = -16.0C at x1 = -1.650m and charge q2 = 38.5nC at the origin (x = 0) must be calculated separately and then summed. The formula used is Fnet = (K * q1 * q3) / D1^2 + (K * q2 * q3) / D2^2, where D1 and D2 are the distances from charge 3 to charges 1 and 2, respectively. The correct net force calculation yields a value of -6.64 * 10^-6 N, indicating the net force acts to the left.

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  • Basic algebra for manipulating equations
  • Knowledge of electric field concepts
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s.bala
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i keep getting this question wrong can some1 help

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two point charges located on the x axis: one charge, q1 = -16.0 , is located at x1= -1.650m ; the second charge, q2= 38.5nC , is at the origin x=0. A
3rd charge is at x=-1.080m and its charge is 45.5nC
wat is the netforce of the charges acting on charge3?
(E_o=8.854*10^-12 therefore K=8.99*10^9)

i know net force on 3 would be to the left and all you have to do is use faradays law twice( Force 1 on 3 and Force 2 on 3) and add those up to get the net force. I keep getting -6.64*10^-6 but its wrong.. can some1help please

Fnet= Force 1 on 3+ Force 2 on 3
=(Kq_1q_3)/D^2(fromx1 to x3) + (Kq_2q_3)D^2(from x2 to x3)

and yes i converted nC int C (factor of 1*10^-9 rii?)
 
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