Find the Distance and Sign of Charge Q for F1 = F2

[Kathi201]

Charge 1 is located along the x-axis and has a magnitude of +1.6E-6C and Charge 2 is located 1m away from charge 1 with a magnitude of -4.8E-6C. A third point charge of magnitude Q is brought from far apart and placed on the line joining the two charges and at a point where the charge q experiences no net force at all. What is the distance of that point from the origin and what should be the sign of this unknown charge q?

This is what I came up with so far...

F1 = K (Q1)(Q2)/X^2
F2 = K (Q2)(Q3)/(1m-x)^2
You set them equal to each other and get
K(Q1)(Q2)/X^2 = K(Q2)(Q3)/(1.00-x)^2
Cancel out K and Q3 leaving;
(1.00-x^2)/X^2 = (Q2)/(Q1)
1.00-x/x = square root (4.8E-6C/1.6E-6C)
1.00-1x = 1.73x
1.00=1.73x +1x
1.00=2.73x
x = .366

Does anybody know if this is correct?

Also I do not know how to figure out what the sign of this charge should be so if anybody could help me I would appreciate it!

 

[LowlyPion]

Charge 1 is located along the x-axis and has a magnitude of +1.6E-6C and Charge 2 is located 1m away from charge 1 with a magnitude of -4.8E-6C. A third point charge of magnitude Q is brought from far apart and placed on the line joining the two charges and at a point where the charge q experiences no net force at all. What is the distance of that point from the origin and what should be the sign of this unknown charge q?

This is what I came up with so far...

F1 = K (Q1)(Q2)/X^2
F2 = K (Q2)(Q3)/(1m-x)^2
You set them equal to each other and get
K(Q1)(Q2)/X^2 = K(Q2)(Q3)/(1.00-x)^2
Cancel out K and Q3 leaving;
(1.00-x^2)/X^2 = (Q2)/(Q1)
1.00-x/x = square root (4.8E-6C/1.6E-6C)
1.00-1x = 1.73x
1.00=1.73x +1x
1.00=2.73x
x = .366

Does anybody know if this is correct?

Also I do not know how to figure out what the sign of this charge should be so if anybody could help me I would appreciate it!

You have chosen a point between the positive and negative charges. There is an E-field between those two charges at all points. What you found is the point at which the magnitudes are equal, but the direction is the same direction for both.

Choose a point outside the 2 charges, away from the larger one and closer to the smaller to fgure your Null point.

 

[Kathi201]

So if I choose a point to the right of the -4.8 charge then the vectors are in the opposite direction. So are you saying in that equation I should add x
so it would be

1.00+x = 1.73x
1.00=1.73x - 1x
1.00 = .73x

x = 1.4m

If that is right then how do I figure out the sign of the charge?

Thanks for your help!

 

[LowlyPion]

So if I choose a point to the right of the -4.8 charge then the vectors are in the opposite direction. So are you saying in that equation I should add x
so it would be

1.00+x = 1.73x
1.00=1.73x - 1x
1.00 = .73x

x = 1.4m

If that is right then how do I figure out the sign of the charge?

Thanks for your help!

Refresh my memory, does an e-field of 0 exert a force on any charge?