Equilateral triangle electrostatics problem

In summary: I'm an idiot i solved it ok so the angle between the two force vectors is 30 degreesok so the law of cosines is C^2 = A^2 + B^2 - 2ABCosCso then C^2 = (7.2)^2 + (10.8)^2 - 2(7.2)(10.8)Cos30C^2 = 115.2 + 116.64 - 43.2Cos30C^2 = 231.84 - 21.6C^2 = 210.24C = 14.5 NIn
  • #1
in10sivkid
36
0
Three point charges of +2μC, +3μC and +4μC are at the vertices of an equilateral triangle ABP, respectively, having sides of 10cm. What is the resulting force R acting on the +4μC charge?


the hint in class is to use the law of cosines

this is how i set it up

force acting on 4uc from 2uc =
(9*10^9 N^2*m^2/C^2)(4uc)(2uc)/(.10m)^2
=7.2 N

force acting on 4uc from 3uc =
(9*10^9 N^2*m^2/C^2)(4uc)(3uc)/(.10m)^2
=10.8 N

now here is where I'm kinda confused i don't entirely see how to use the law of cosines here

from the information i found out that the angle between the resultant of 7.2 and 10.8 = 48.2

by using cos(theta) = 7.2/10.8

then i used C^2 = A^2 + B^2 - 2ABCos(theta)
C^2 = 7.2^2 + 10.8^2 - 2(7.2)(10.8)(Cos48.2)
C= 8.05 N
C= resultant vector

but my teacher says the answer to the problem is 15.7N

what have i done wrong?
 
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  • #2
in10sivkid said:
force acting on 4uc from 2uc =
(9*10^9 N^2*m^2/C^2)(4uc)(2uc)/(.10m)^2
=7.2 N
OK. What direction does it have? (Let A & B be on a horizontal line, with P above it.)

force acting on 4uc from 3uc =
(9*10^9 N^2*m^2/C^2)(4uc)(3uc)/(.10m)^2
=10.8 N
OK. Again, what direction does it have?

now here is where I'm kinda confused i don't entirely see how to use the law of cosines here
To add the two vectors, draw a triangle. Use the law of cosines to calculate the resultant (which is one side of the triangle).

from the information i found out that the angle between the resultant of 7.2 and 10.8 = 48.2
Not sure what you mean, but the angle between those two vectors is not 48.2 degrees. (Draw a picture.)

by using cos(theta) = 7.2/10.8
??

Hint: Find the angle between the two vectors by drawing a careful diagram. That's the angle you will use in the law of cosines. (What are the angles in an equilateral triangle?)
 
  • #3
ok i think i might have conveyed what I've done in the problem badly so i'll explain a bit more thoroughly what i did because there seems to be some confusion on my part and doing this usually helps me think through

well i setup the triangle so that A and P were on the horizontal and P was on top

so A= 2uc
P = 3uc
B= 4uc

ok thus the direction of all the forces acting upon these charges will be in the opposite direction between they are like charges and like charges repel

so A acting on B will push it horizontally away
and P acting on B will push away on a angle

ok so I got the force of A acting on B = 7.2 N
P acting on B = 10.8 N


now Doc Al you highlighted my force calculation of the 3uc acting on the 4uc (which is P acting on B) as possibly incorrect...what am i missing here? i thought the direction was away...so what exactly do i need to fix here

ok so would the angle then that it would make would be a 60 degree angle because of the equilateral triangle?

oh wow...ok i am thinking this through and thus i have made the following changes and i think i got the right answer please tell me if I'm correct

so A acting on B = 7.2 N
P acting on B = -10.8 N

the angle made is 60 degrees

thus using the law of cosines i have as follows

C^2 = (7.2^2) + (-10.8^2) - 2(7.2)(-10.8)Cos60
C^2 = 246,24
C = 15.7 N

well that's the answer that its supposed to be...but please conceptually tell me why the 10.8 is negative and the 7.2 isn't negative...thats the only piece of information i am missing to completely understand this problemd
 
  • #4
in10sivkid said:
so A acting on B will push it horizontally away
and P acting on B will push away on a angle

ok so I got the force of A acting on B = 7.2 N
P acting on B = 10.8 N
OK.


now Doc Al you highlighted my force calculation of the 3uc acting on the 4uc (which is P acting on B) as possibly incorrect...what am i missing here? i thought the direction was away...so what exactly do i need to fix here

ok so would the angle then that it would make would be a 60 degree angle because of the equilateral triangle?
Exactly. The angle of each force is 60 degrees above the horizontal. So what's the angle between those two forces if you treat them as sides of a triangle? The triangle I'm talking about is the triangle you would draw when adding those two forces as vectors. (Put the tail of one on the point of the other.)

oh wow...ok i am thinking this through and thus i have made the following changes and i think i got the right answer please tell me if I'm correct

so A acting on B = 7.2 N
P acting on B = -10.8 N
Where did you get the negative from?

the angle made is 60 degrees

thus using the law of cosines i have as follows

C^2 = (7.2^2) + (-10.8^2) - 2(7.2)(-10.8)Cos60
C^2 = 246,24
C = 15.7 N

well that's the answer that its supposed to be...but please conceptually tell me why the 10.8 is negative and the 7.2 isn't negative...thats the only piece of information i am missing to completely understand this problemd
The 10.8 N force should not be negative. And the angle between the sides of the triangle is not 60 degrees.

Draw the two forces just like you would if you were adding them (which you are!).
 
  • #5
oh man...im sorry I'm now confused i thought I had it right...

so let me get this straight i have the two force vectors right of 7.2 and 10.8

and they are the side of a new triangle...and the other line that connects those two force vectors together will be the resultant that I'm looking for ..right?

ok would the angle between the two lines then be 30 degrees?

do I have the law of cosines right?

how am i supposed to plug in the values?
 
  • #6
ok so i have 7.2 and 10.8 which trigometric function am i supposed to use here
 
  • #7
ok i did the law of cosines knowing that the answer is supposed to 15.7 N and i got the angle to be 120 degrees...is that possible?

are you sure the angle between 7.2 and 10.8 isn't supposed to be 60, because the way i have it made up the 120 degree angle is the angle that is made between the two triangles

man I am just getting confused.
 
  • #8
in10sivkid said:
ok i did the law of cosines knowing that the answer is supposed to 15.7 N and i got the angle to be 120 degrees...is that possible?
Did you actually draw a diagram like I asked? One vector makes an angle of 60 degrees with the horizontal. Now when you add the second vector, which also makes an angle of 60 degrees to the horizontal but slants in the opposite direction, the "triangle" formed has 120 degrees between the sides. (The sides of that triangle are the two vectors you are adding.) You must draw this. Do not settle for a lucky guess.
 
  • #9
yes i did draw it but i think my problem is that I'm not good at drawing vectors...thats my problem the way i did it the angle looked like it was 60 degrees
 
  • #10
Law of Cosines would lead to many more mistakes since you don't understand vector concepts here - so try to get it working piece by piece and put it all together algebraically
 
Last edited:
  • #11
in10sivkid said:
yes i did draw it but i think my problem is that I'm not good at drawing vectors...thats my problem the way i did it the angle looked like it was 60 degrees
I suspect that something is wrong with how you are drawing your diagram. In any case, rather than going by what the angle on your diagram looks like, use geometry to figure out what the angle must be.

I drew a quick (not to scale) diagram showing the triangle formed when adding two vectors:
[tex]\vec{R} = \vec{A} + \vec{B}[/tex]

You are trying to find the resultant, which is side "R" of the triangle. Convince yourself that the angle opposite that side--which is the angle used in the law of cosines--equals 60 + 60 = 120 degrees.

(Using the law of cosines is only one way of finding the resultant. But since your instructor brought it up, you may as well learn it.)
 

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FAQ: Equilateral triangle electrostatics problem

1. What is an equilateral triangle electrostatics problem?

An equilateral triangle electrostatics problem is a type of physics problem that involves calculating the electric field and potential at various points within and around an equilateral triangle. This type of problem typically involves applying the principles of electrostatics, such as Coulomb's law and the superposition principle, to determine the behavior of electric charges in this specific geometric configuration.

2. How do I approach solving an equilateral triangle electrostatics problem?

First, draw a diagram of the equilateral triangle and label the known values, such as the distances between charges and the magnitude of the charges. Then, use Coulomb's law to calculate the electric field and potential at each point. Finally, apply the superposition principle to find the total electric field and potential at each point.

3. What are some common mistakes when solving an equilateral triangle electrostatics problem?

Some common mistakes include forgetting to take into account the direction of the electric field or potential, not properly applying the superposition principle, and using incorrect values for the distance or magnitude of charges. It is important to double-check all calculations and make sure they are consistent with the principles of electrostatics.

4. Can I use different units when solving an equilateral triangle electrostatics problem?

Yes, as long as the units are consistent. For example, you can use meters for distances and Coulombs for charges. However, it is important to keep track of units and convert them if necessary to ensure accurate calculations.

5. How can I check my answer for an equilateral triangle electrostatics problem?

You can check your answer by using the equations for electric field and potential to calculate the values at each point and comparing them to your calculated values. Additionally, you can use the symmetry of the equilateral triangle to check if your answer makes sense, such as if the electric field is zero at the center of the triangle.

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