Which Arrangement of Charged Balls Results in the Largest Net Force?

[candyq27]

Homework Statement

Three balls, with charges of +4q, -2q, -q, are equally spaced along a line. The spacing between the balls is r. We can arrange the balls in three different ways: (1) +4q, -2q, -q; (2) -2q, +4q, -q; (3) +4q, -q, -2q.
(a) Which ball experiences the largest magnitude net force in each of the three cases?
(b) Express your answers below in terms of k, Q, and r. Calculate the magnitude and direction of the force applied to the ball that has the -q charge in each of the three cases.
(c) In which case does the ball of charge +4q experience the largest force?

Homework Equations

k=9x10^9
magnitude of the force between charges: F=k(q1)(q2)/(r)^2 and F=qE
magnitude of the electric field: E=kq/r^2

The Attempt at a Solution

(a) Case 1: The -2q charge has the largest force because the -q charge experiences a large force to the right and a small to the left, the -2q charge experiences a large force to the right and a large force to the right, and the +4q charge has a large force to the right and a small force to the right.
Case 2: The -q charge and the -2q charge have equal magnitude but opposite direction net forces.
Case 3: The -q charge has the largest force because the -2q charge has a large force to the left and a small force to the right and the +4q has a large force to the left and a small force to the left.
I'm just learning how to do this so I'm not sure if these are correct. Please let me know so I can move on to part (b). Thanks!

 

[candyq27]

Can anyone tell me if this is correct? I posted a few days ago but no one has responded.

 

[Warr]

the first 2 cases are correct. The third one isn't.

The -q feels a repulsive force of magnitude 2 in the -x direction due to the -2q , and a attractive force of magnitude 4 in the -x direction due to the +4q: total magnitude = 6 in -ve x direction.

The +4q feels a an attractive force of magnitude 4 in the +x direction due to the -q charge and an attractive force of 2 (2*4/(2)^2=2) in the +ve x direction due to the -2q charge. Total magnitude 4+2=6.

Hence the +4q and -q feel the same magnitude of force, in opposite directions.

(Note, these aren't the absolute magnitudes, just the relative magnitudes.)

 

[candyq27]

Ok thanks
So for part (c), the ball with the +4q charge would experience the largest force in Case 1 because of the large attraction to -2q and the smaller attraction to -q.
Now for part (b) I know I use the equation F=k(q1)(q2)/r^2, so do i do it for each of the surrounding balls and then add them together to get the net force? Thanks!

 

[Warr]

That is correct

 

[candyq27]

yess thank you

I think I also figured out part (b) but it is a lot to write out and I don't know how to make the formulas look nice so I'm not sure if I should even try to write them down so if you don't understand how I'm writing this I apologize...
Case 1: F= -[(k*2q*q)/(r)^2] + [(k*4q*q)/(2r)^2]
Case 2: F= +[(k*4q*q)/(r)^2] - [(k*2q*q)/(2r)^2]
Case 3: F= +[(k*4q*q)/(r)^2] - [(k*2q*q)/(r)^2]
Hopefully that's right!