Force on a Charged Object in a Field

In summary, the problem involves a charge of +7.55 μC located in an electric field with x and y components of Ex = 6.14E+3 N/C and Ey = 8.17E+3 N/C, respectively. The question is asking for the magnitude of the force on the charge, which can be found using F=qE and the parallelogram method to get the magnitude of the vector force. The initial approach of finding the average of the electric fields and dividing by q is off track and a diagram can be helpful in solving the problem. It is also important to understand the relationship between the magnitude of the force and the electric field, as well as how to calculate the magnitude of a vector given
  • #1
Boozehound
29
0
A charge of q = +7.55 μC is located in an electric field. The x and y components of the electric field at this point are Ex = 6.14E+3 N/C and Ey = 8.17E+3 N/C, respectively. What is the magnitude of the force on the charge?

i know for sure that i am going in the wrong direction in this one. if i could just get some guidance as to what to do that would be a big help.

i started off by taking the two electric fields and adding them together and dividing by two to get an average...why? i don't know. then i took that answer and divided it by (q). the closest equation that i found in my book and my notes is no where near what i need to figure out this problem. and I've tried to look for multiple equations that i could use together but no dice. thank you for your time.
 
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  • #2
Okay, first of all draw a picture of the forces acting on it. Get a grip or an idea of how it's happening. It's a lot easier to really see these problems when you can see how it's happening. Remember when you did vectors a long time ago in Physics =p. Same ideas apply.

Remember also, F = qE.
 
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  • #3
since F ,like E ,have two components, Fx and Fy,get them use F=q*E;

//the q is the same//

then use parallelogram method to get the magnitude of the F,because it is vector
 
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  • #4
Boozehound said:
i started off by taking the two electric fields and adding them together and dividing by two to get an average...why? i don't know.

Yeah, that's getting off track right from the start. You do understand that the electric field is a vector quantity, right? The electric field that they have given you is not pointing in just the x direction, or just the y direction. It is point in a direction that can be represented by a sum of x and y components. This is a vector sum.

Now, the question asks what the magnitude of the force on the particle is. So you have to answer two questions:

1. How is the magnitude of the force related to the magnitude of the electric field? Your book definitely has the answer, although it's something you should know anyway, because it comes right from definition of the electric field.

2. How do I calculate the magnitude of the electric field? This question boils down to, how do I calculate the magnitude of any vector, if its x and y components have been given?

Drawing a diagram would be very helpful.
 
  • #5
thank you...just that little formula helped me. haha i found it in my book but i have been staring at this problem for so long that it never occurred to me to flip around the equation...wow
 
  • #6
enricfemi said:
Fx=q*Ex;
Fy=q*Ey;//the q is the same//

then use parallelogram method to get the magnitude of the F,because it is vector

You've basically given up most of the solution. Don't do that from now on. You're not doing this guy any favours by solving his problem for him.
 
  • #7
so sorry, when i replied,i didn't see any answer in this thread.AngeloG 's reply will be ok.
 
  • #8
cepheid said:
You've basically given up most of the solution. Don't do that from now on. You're not doing this guy any favours by solving his problem for him.

ok,i will edit it immediately
 
  • #9
enricfemi said:
ok,i will edit it immediately

Ha, it's ok. I'm not a moderator, so I have no authority here. I didn't mean to be harsh. You did help the guy, after all (and editing it would serve no purpose because he has alread read it). It's just that there is a general rule at this forum to help people to arrive at the answers for themselves while we give guidance, directions, and hints, so that they learn something from doing their homework (which is the whole point of it! :smile: ). In contrast, they won't learn much if we solve their homework for them. I just though I'd let you know for future reference, so that when the answer is sitting there right before their eyes in the textbook, they will learn to look for it. No hard feelings, I hope.
 
  • #10
cepheid said:
Ha, it's ok. I'm not a moderator, so I have no authority here. I didn't mean to be harsh. You did help the guy, after all (and editing it would serve no purpose because he has alread read it). It's just that there is a general rule at this forum to help people to arrive at the answers for themselves while we give guidance, directions, and hints, so that they learn something from doing their homework (which is the whole point of it! :smile: ). In contrast, they won't learn much if we solve their homework for them. I just though I'd let you know for future reference, so that when the answer is sitting there right before their eyes in the textbook, they will learn to look for it. No hard feelings, I hope.

:!) :!) :!) :!) :!) :!) :!) :!)
of couse i feel no hard or unhappy.i just like doing things immediately,since i know you are right.
:rofl:
 
  • #11
I have a similar question, and I'm getting confused on the tension part of the forces. Would the 3 forces be the horiz/vert of the tension force, then the electric field force? So you get Tcos(theta) + mg-Tsin(theta) + Eq = 0 ...then solve for q? I'm pretty sure this is wrong...any guidance?
 
  • #12
ooops I posted on the wrong problem I think haha
 

1. What is the formula for calculating the force on a charged object in a field?

The formula for calculating the force on a charged object in a field is F = qE, where F represents the force, q is the charge of the object, and E is the strength of the electric field.

2. How does the direction of the electric field affect the force on a charged object?

The direction of the electric field determines the direction of the force on a charged object. The force will act in the same direction as the electric field if the charge on the object is positive, and in the opposite direction if the charge is negative.

3. What happens to the force on a charged object if the charge or the electric field changes?

If the charge or the electric field changes, the force on the charged object will also change. As the charge increases, the force will increase as well, and vice versa. Similarly, if the electric field becomes stronger, the force will become stronger, and if it becomes weaker, the force will also weaken.

4. Can there be a force on an uncharged object in an electric field?

No, an uncharged object will not experience a force in an electric field. This is because the force on a charged object in a field is directly proportional to the charge of the object, and an uncharged object has a charge of zero.

5. How does the distance between the charged object and the source of the electric field impact the force?

The force on a charged object in an electric field decreases as the distance between the object and the source of the field increases. This is because the strength of the electric field decreases with distance, leading to a weaker force acting on the charged object.

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