Resultant Force on Charge in the middle of square

[katia11]

Homework Statement

(a) Find the resultant force on a negative charge of -3q, placed at the center of a square of side b, which has charges on the 4 corners, respectively, q, 2q, -4q, 2q in clockwise order. (Magnitude and direction)

Homework Equations

F= k*q1*q*(1/ r^2)

Ftotal= F₁₅ + F₂₅ + F₃₅ + F₄₅

The Attempt at a Solution

I drew the figure, and see that r = square root 2 *b* (1/2)

(pythagorean theorem)

One question is that F25 and F45 are equal and opposite. Do they cancel each other out? If so, I wouldn't have to calculate the forces for those, yes?

I'm just a little confused. Sorry if it's super easy.

 

[Andrew Mason]

Homework Statement

(a) Find the resultant force on a negative charge of -3q, placed at the center of a square of side b, which has charges on the 4 corners, respectively, q, 2q, -4q, 2q in clockwise order. (Magnitude and direction)

Homework Equations

F= k*q1*q*(1/ r^2)

Ftotal= F₁₅ + F₂₅ + F₃₅ + F₄₅

The Attempt at a Solution

I drew the figure, and see that r = square root 2 *b* (1/2)

I assume you mean:
r = \sqrt{2}b/2 = .707b

One question is that F25 and F45 are equal and opposite. Do they cancel each other out? If so, I wouldn't have to calculate the forces for those, yes?

Correct. Just draw each force vector and add them together (tail to head). You will see that the net force is in the direction of q.

AM

 

[katia11]

Ok awesome!

So now I have
F₁₅= (2ke₃q²)/(b²) N

and
F₃₅=(2ke₄q²)/(b²) N

But how do I find the x and y components without numbers? Do I just put in

F_15x= F₁₅ sin(135)
F_15y= F₁₅ cos(135)

and

F_35x= F₃₅ sin(135)
F_35y= F₃₅ cos(135)

The fact that there's no value for q shouldn't be throwing me off this much, but it is.

 

[Andrew Mason]

Ok awesome!

So now I have
F₁₅= (2ke₃q²)/(b²) N

and
F₃₅=(2ke₄q²)/(b²) N

But how do I find the x and y components without numbers? Do I just put in

F_15x= F₁₅ sin(135)
F_15y= F₁₅ cos(135)

and

F_35x= F₃₅ sin(135)
F_35y= F₃₅ cos(135)

The fact that there's no value for q shouldn't be throwing me off this much, but it is.

Think of the -4q charge as made up of two charges, +q and -5q. Draw the force vector resulting from each charge. The force vector from the +q charge cancels that from the opposite corner. You are then left with a single force vector from the -5q charge that is a distance .707b away. Apply Coulomb's law to determine the magnitude of the force. The direction is obvious. What is the angle between the force vector and one of the sides?

AM

 

[katia11]

45 degrees?

I think I was confused because I made the middle charge the origin and thought the angle came from that. . .

And I don't really understand why we split the -4q because if one if +q and one is -5q we would get a net result of -4q anyway. . .

I assume the direction is towards the first corner q.

Thanks so much again, sorry if something is going over my head. . .