Posted by Anonymous on .
In the figure below, four particles are fixed along an x axis, separated by distances d = 4.00 cm. The charges are q1 = +2e, q2 = e, q3 = +e, and q4 = +5e, with e = 1.60 × 1019 C. What is the value of the net electrostatic force on (a) particle 1 and (b) particle 2 due to the other particles?
q1 is at the origin and the rest of the particles are to the right a distance d from one another. therefore the order of particles is q1,q2,q3,q4 each with a distance d in between.
Got part a to be 5.596E26 but i don't know how to do b. please help

physicshelp 
bobpursley,
The only difference in part 2 from part 1 is that you have to consider space and direction of forces. Do this, sketch which way the forces are.
at q2, reverse the sign of F since it is on the left side.
Ftotal2= k/d^2 (q1q2+q3q2+q4q2/2^2 )
That should do it.
If you want to the direction, You need to do the vector form of the equation. 
physicshelp 
Anonymous,
i ended up with 2.8767E29 but its incorrect...i don't understand what im doing wrong still

physicshelp 
sabrina,
5. If the voltage between the two plates of an electric air cleaner is 500 V, how fast would a 1012 kg soot particle with 1 „e 1011 C of charge on it be moving if it went from the negative plate to the positive?