# Magnitude and Direction of the Force on a Charge

• Physistory
I am glad I could be of assistance! In summary, the conversation involved finding the force F (a vector) on a -10 nC charge in a figure consisting of a rectangle with various charges. The distance between the top charges was 3.0 cm and the distance between the top and bottom right charges was 1.0 cm. Using the Law of Sines and Pythagoras' equation, the magnitude and direction of the net force were found to be 4.3*10^-3 N at 253° counterclockwise from the +x-axis.
Physistory

## Homework Statement

What is the force F (a vector) on the -10 nC charge in the figure? Give your answer as a magnitude and an angle measured cw or ccw (specify which) from the +x-axis.

The figure shown is in the shape of a rectangle. On the top left is a +15 nC charge; on the top right is a -5.0 nC charge; on the bottom right is a -10 nC charge. There is no charge on the bottom left. The distance between the top charges is 3.0 cm. The distance between the top and bottom right charges is 1.0 cm.

## Homework Equations

Fnet = F1on3 + F2on3 = K|q1||q2|/r2
a2=b2+c2

a/sin(A)=b/sin(B)

K ≅ 9.0*10^9 N*m^2/C^2

## The Attempt at a Solution

The first thing I did was write out the first equation and plug in the given values as (9*10^9)(15*10^-9)(10*1-^-9)/(.0316)2 for F1on3. I found .0316 by using Pythagorus' equation to find the distance between the first and third charges. Then I plugged in (9*10^9)(5*10^-9)(10*10^-9)/(.01)2 for F2on3. Typing these values into my calculator, I got 1.35*10^-3 N and 4.5*10^-3 N. This is where I first got stuck, because I am figuring that one value is supposed to be subtracted from the other, because they would be of differently directed charges. 1on3 would be directed southeast and 2on3 would be directed straight down.

I tried to get closer to the correct answer by using the Law of Sines to find the angle between the first charge and the 3 cm distance. I wrote out 3.16/sin(90) = 1/sin(x). For x, I got 18.4°. I plugged this into the vertical component of F1on3 to get 4.26*10^-4 N. 2on3 points straight down, so I opted to subtract 4.26*10^-4 N from that. I ended up with 4.07*10^-3 N, which is close to the correct magnitude, but not quite.

The correct answer provided in the back of my textbook is 4.3*10^-3 N, at 253° ccw. As far as correctly reaching the magnitude, I'm confused, and as far as how to obtain the right direction, I am stumped. Any and all hints, tips and suggestions would be greatly appreciated.

Additionally, this is my first post, so please forgive me if my formatting is less than ideal.

Physistory said:
2on3 points straight down, so I opted to subtract 4.26*10^-4 N from that. I ended up with 4.07*10^-3 N
Good, but you now need to combine that with the x component of 1on3 to get the overall force. Use Pythagoras again.
(I get a bit more than 4.3*10-3.)

haruspex said:
Good, but you now need to combine that with the x component of 1on3 to get the overall force. Use Pythagoras again.
(I get a bit more than 4.3*10-3.)

Thank you for the response! My attempt is as follows:

F1on3xcos(18.4) = (1.35*10^-3)cos(18.4) = 1.28*10^-3 N
This component can be inserted either at the tip or the tail of the y-component of 2on3, so I formed a triangle whose hypotenuse is the overall force. Due to complementary angles, I used 18.4° again. Now, my final equation is as follows:

Fnet = √(4.07*10^-3)2 + (1.28*10^-3)2 = √(1.65*10^-5 + 1.63*10^-6) = √1.81*10^-5 = 4.25*10^-3 N, which approximates to 4.3*10^-3. I suppose my answer is a bit smaller due to slight differences in rounding in previous steps. Or I am just overlooking something crucial.

I think I have some insight now regarding the direction. 90° + 18.4° = 108.4°, which subtracted from 360° is 251.6° which is very close to my textbook's answer of 253°. The direction is counterclockwise because the net force is southeast and 1on3 is significantly larger than 2on3, thus initially pushing the force to the left and resulting in a counterclockwise southeast force overall. Have I improved, or is there something else I am forgetting to do? Thanks again!

Physistory said:
4.25*10^-3 N, which approximates to 4.3*10^-3
Good enough.
Physistory said:
90° + 18.4° =
But 18.4 degrees is the angle of the 1on3 force, not the angle of the net force.
Physistory said:
I formed a triangle whose hypotenuse is the overall force.
Can you find the angle of that hypotenuse to the x axis?

haruspex said:
But 18.4 degrees is the angle of the 1on3 force, not the angle of the net force.

Oh, of course. Little mistakes like that tend to stump me! I see now that I used the wrong angle there because the 1on3 force and the net force point in similar directions. Thank you.

haruspex said:
Can you find the angle of that hypotenuse to the x axis?

I believe I found the correct angle using the Law of Sines, as follows:

(4.3*10^-3)/sin(90) = (1.28*10^-3)/sin(x)
sin(x) = (4.3*10^-3)/(1.28*10^-3) = 0.297...
sin^-1(0.297) = 17.3°

90° + 17.3° = 107.3°
360° - 107.3° = 252.7° ≅ 253°, counterclockwise for the reasons I postulated in my previous response.

Is this correct?

Physistory said:
Oh, of course. Little mistakes like that tend to stump me! I see now that I used the wrong angle there because the 1on3 force and the net force point in similar directions. Thank you.
I believe I found the correct angle using the Law of Sines, as follows:

(4.3*10^-3)/sin(90) = (1.28*10^-3)/sin(x)
sin(x) = (4.3*10^-3)/(1.28*10^-3) = 0.297...
sin^-1(0.297) = 17.3°

90° + 17.3° = 107.3°
360° - 107.3° = 252.7° ≅ 253°, counterclockwise for the reasons I postulated in my previous response.

Is this correct?
Looks right.

haruspex said:
Looks right.

Thank you very much! Your hints have been a huge help.

## What is the magnitude of the force on a charge?

The magnitude of the force on a charge is determined by the Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

## How is the direction of the force on a charge determined?

The direction of the force on a charge can be determined by using the right-hand rule. This rule states that if you point your right thumb in the direction of the moving charge and your fingers in the direction of the magnetic field, then the direction of the force will be perpendicular to both your thumb and fingers.

## Can the direction of the force on a charge be changed?

Yes, the direction of the force on a charge can be changed by either changing the direction of the magnetic field or the direction of the charge's motion. The force will always be perpendicular to both the magnetic field and the charge's velocity.

## What factors affect the magnitude of the force on a charge?

The magnitude of the force on a charge is affected by the strength of the magnetic field, the amount of charge, and the velocity of the charge. Increasing any of these factors will result in a stronger force on the charge.

## How is the force on a charge related to its motion?

The force on a charge is directly related to its motion. If the charge is moving parallel to the magnetic field, there will be no force on it. However, if the charge is moving perpendicular to the magnetic field, the force will be at its maximum strength.

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