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Resultant Force on Charge in the middle of square

by katia11
Tags: charge, force, middle, resultant, square
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katia11
#1
Jan8-11, 01:29 PM
P: 18
1. The problem statement, all variables and given/known data
(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)


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

Ftotal= F15 + F25 + F35 + F45

3. 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 eachother 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.
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Andrew Mason
#2
Jan8-11, 02:38 PM
Sci Advisor
HW Helper
P: 6,654
Quote Quote by katia11 View Post
1. The problem statement, all variables and given/known data
(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)


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

Ftotal= F15 + F25 + F35 + F45

3. The attempt at a solution
I drew the figure, and see that r = square root 2 *b* (1/2)
I assume you mean:
[tex]r = \sqrt{2}b/2 = .707b[/tex]
One question is that F25 and F45 are equal and opposite. Do they cancel eachother 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
#3
Jan8-11, 03:46 PM
P: 18
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

F15x= F15 sin(135)
F15y= F15 cos(135)

and

F35x= F35 sin(135)
F35y= F35 cos(135)

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

Andrew Mason
#4
Jan8-11, 08:24 PM
Sci Advisor
HW Helper
P: 6,654
Resultant Force on Charge in the middle of square

Quote Quote by katia11 View Post
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

F15x= F15 sin(135)
F15y= F15 cos(135)

and

F35x= F35 sin(135)
F35y= F35 cos(135)

The fact that theres 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
#5
Jan8-11, 09:12 PM
P: 18
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. . .


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