Electric Charge Problem: Determining Resultant Electrostatic Force"

[frankfjf]

I got this problem wrong in my text's student companion website, but I don't understand why. Could someone let me know what went wrong?

Homework Statement

What are the horizontal and the vertical components of the resultant electrostatic force on the charge in the lower left corner of the square if q = 0.10 micro Coulombs (Or uC as I cannot remember how to type the Greek letter equivalent.) and a = 5.0cm?

Homework Equations

Fh = F-q + F-2q(cos(45))

Fv = -F+q + F-q(sin(45))

The Attempt at a Solution

I get .17E-6 N for Fh and -2.1E-7 N for Fv. I think I may have mixed up some decimal places on Fh, but the solution presented is .17 N for Fh and -0.046 for Fv. I just don't understand what I did wrong though.

 

[mbrmbrg]

for Fh: your first term should be $$F_{-2q}$$, and your second term should be $$F_{-q}\cos(45^o)$$

for both: Please show more work. What is the general formula for the electrostatic force? What values did you use for each variable in the equation?

 

[frankfjf]

My apologies. I'll try out your suggestion for Fh. Otherwise...

Fh = ((8.99E9)(2E-7)(1E-7)/.05^2) + (((8.99E9)(2E-7)^2)/.05^2)cos(45)

Fv = -((8.99E9)(2E-6)(1E-6)/.05^2) + ((8.99E9)(2E-6)(1E-6)/.05^2)sin(45)

 

[mbrmbrg]

For Fv: as in Fh, the charges should be multiplied by 10^-7

 

[frankfjf]

After your suggestions I get .19E-8 for FH and -.02E-8. Why do my answers still differ from what the website presents? I don't get it.

 

[mbrmbrg]

My apologies. I'll try out your suggestion for Fh. Otherwise...

Fh = ((8.99E9)(2E-7)(1E-7)/.05^2) + (((8.99E9)(2E-7)^2)/.05^2)cos(45)

Fv = -((8.99E9)(2E-6)(1E-6)/.05^2) + ((8.99E9)(2E-6)(1E-6)/.05^2)sin(45)

for the charge on the far corner: the distance is not 0.05m. It is along the diagonal of the square, not along one of the sides.
I'm checking now to see if that clears things up...

edit: I got the correct answer for both Fh and Fv when I took both of my suggestions (and the rest of your work ).

 

[frankfjf]

So for the charge in the far corner, it's not 0.05m? If not then how can I get the right measurement? The diagram doesn't give any indication of what the diagonal is. Or do you mean just the two suggestions you gave and the diagonal is a red herring so to speak?

 

[mbrmbrg]

Cheese, frank, I wouldn't trick you like that!

You don't know the formula for the diagonal of a square? It's the same as the hypotenuse of a 45-45-90 triangle.

 

[frankfjf]

Ah okay, that helped. Thanks!