Magnitude and Direction of the Force on a Charge

  • #1
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.
 

Answers and Replies

  • #2
haruspex
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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.)
 
  • #3
Physistory
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!
 
  • #4
haruspex
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4.25*10^-3 N, which approximates to 4.3*10^-3
Good enough.
90° + 18.4° =
But 18.4 degrees is the angle of the 1on3 force, not the angle of the net force.
I formed a triangle whose hypotenuse is the overall force.
Can you find the angle of that hypotenuse to the x axis?
 
  • #5
Physistory
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.

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?
 
  • #6
haruspex
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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.
 
  • #7
Physistory
Looks right.

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

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