# Net force on a charge in a square

## Homework Statement

There are 4 charges in the corner of a square with the lengths of the sides measuring 18.2cm
q1= 2.34μC (top left) q2=-1.58μC (top right) q3= 2.03μC (bottom left) q4= -3.00μC (bottom right)

→Calculate net force on q4
→Calculate net force at the center of the square

## Homework Equations

Fnet=|KQ|/r^2
K constant = 8.99x10^9

## The Attempt at a Solution

q1= [(8.99x10^9)(2.34x10^-6)]/(0.182)^2 = 635086.3422 @ 315°
q2= [(8.99x10^9)(1.58x10^-6)]/(0.182)^2 = 428818.9832 @270°
q3= [(8.99x10^9)(2.03x10^-6)]/(0.182)^2 = 550950.9721 @ 0°

for my next step I was going to calculate x and y components and then plug it into the
d=*sqrt [(x^2)+(y^2)]

to calculate net force on q4... but I'm very uncertain about how to continue or if my previous calculations are even right.

Related Introductory Physics Homework Help News on Phys.org
You're on the right track. If you calculate the components for each, then you can find the magnitude from the addition of all those components.

Remember to watch your signs on the components too. Also, for the calculation of the force of q1, there's an error in the value for the separation -- you want to use the length of a diagonal instead of a side length.

• 1 person
You're on the right track. If you calculate the components for each, then you can find the magnitude from the addition of all those components.

Remember to watch your signs on the components too. Also, for the calculation of the force of q1, there's an error in the value for the separation -- you want to use the length of a diagonal instead of a side length.
do I square the distance anyway?

Yes, you want to square the distance, but use the length of the diagonal for the force from q1.

• 1 person
Solution??

## Homework Statement

There are 4 charges in the corner of a square with the lengths of the sides measuring 18.2cm
q1= 2.34μC (top left) q2=-1.58μC (top right) q3= 2.03μC (bottom left) q4= -3.00μC (bottom right)

→Calculate net force on q4
→Calculate net force at the center of the square

## Homework Equations

Fnet=|KQ|/r^2
K constant = 8.99x10^9

## The Attempt at a Solution

q1= [(8.99x10^9)(2.34x10^-6)]/(0.182)^2 = 635086.3422 @ 315°
q2= [(8.99x10^9)(1.58x10^-6)]/(0.182)^2 = 428818.9832 @270°
q3= [(8.99x10^9)(2.03x10^-6)]/(0.182)^2 = 550950.9721 @ 0°

for my next step I was going to calculate x and y components and then plug it into the
d=*sqrt [(x^2)+(y^2)]

to calculate net force on q4... but I'm very uncertain about how to continue or if my previous calculations are even right.
Ok so this is what I did
→ Using pythagoreon theorem I found the length of the diagonal line toward q4 which = 025738684 m

I applied this to the previous calculation that I did wrong. But also, wouldn't the angle of q2 be 90° because it is a negative force and therefore repels away from the negative force of q4?

Anyway this is what I did
q1= [(8.99x10^9)(2.34x10^-6)]/(0.25738684)^2 = 317543.2411 @ 315°
q2= [(8.99x10^9)(1.58x10^-6)]/(0.182)^2 = 428818.9832 @90°
q3= [(8.99x10^9)(2.03x10^-6)]/(0.182)^2 = 550950.9721 @ 0°

X and Y components:
E1 -- x: 224536.9791 y: -224536.9791
E2 -- x: 0 y: 428818.9832
E3 -- x: 55090.9721 y: 0

I then added the x components and the y components to calculate E4
x: 775487.9512 y: 204282.0041

I then used the distance formula to calculate the net field on E4:
FnetE4: *sqrt [(775487.9512^2)+(204282.0041^2)]
FnetE4: 801943.0776C ???

Do I still need to calculate the angle?
I did but it doesn't seem right.

Angle tan -1:y/x
=204282.0041/775487.9512
=14.8°

That all looks right for the most part, but because you calculated the net field due to all the charges, you have to multiply by q4 to get the net force.

• 1 person
haruspex
Homework Helper
Gold Member
2020 Award
I then added the x components and the y components to calculate E4
x: 775487.9512 y: 204282.0041
You mean, the field at q4 due to the other three charges, right?
I then used the distance formula to calculate the net field on E4:
FnetE4: *sqrt [(775487.9512^2)+(204282.0041^2)]
FnetE4: 801943.0776C ???
Distance formula? C?
You are asked for the force on q4, but it looks to me that you have only got as far as the field.
Do I still need to calculate the angle?
I did but it doesn't seem right.

Angle tan -1:y/x
=204282.0041/775487.9512
=14.8°
Why does that seem wrong?

You mean, the field at q4 due to the other three charges, right?
Distance formula? C?
You are asked for the force on q4, but it looks to me that you have only got as far as the field.

Why does that seem wrong?
Yes the charge on q4 due to the other 3.
So I would use F=qE to calculate the net force on it right?
It just seems wrong because the direction would be north east?

So then to calculate the net force at the center of the square I've taken half of the distance I had for the charge of q1 -> q4 = 25.738684 / 2 = 12.869342cm

So to calculate the center force I used the formula

Fnet= |Kq1q2|/r2 once for the left side and once for the right side of the square

so Fnet on the left side was =0.042704298 and =.0426126

I am a little lost how to continue or if that is even right... Any help is appreciated!

Yes, you have to do F = qE to get the net force. Also keep in mind that since q4 is negative, the force will be in the opposite direction as the field, so the force will actually be southeast.

And for the center of the square, it doesn't make since to calculate the force at the center of it, since there's nothing in the center to have a force exerted upon it. I assume it means field, so use the field equation. And I don't really understand your process. What do you mean by the left half and the right half of the square? Shouldn't you calculate the fields from each of the four charges and then add them up?

haruspex