Force on a Charged Object in a Field

AI Thread Summary
A charge of +7.55 μC in an electric field with components Ex = 6.14E+3 N/C and Ey = 8.17E+3 N/C requires calculating the force using the formula F = qE. The discussion emphasizes the importance of treating electric fields as vector quantities, combining their x and y components to find the resultant force. Participants suggest drawing diagrams to visualize the problem and using the parallelogram method for vector addition. The conversation highlights a common mistake of averaging electric field components instead of summing them as vectors. Overall, the focus is on guiding problem-solving techniques rather than providing direct answers.
Boozehound
Messages
29
Reaction score
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.
 
Physics news on Phys.org
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.
 
Last edited:
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
 
Last edited:
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.
 
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
 
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.
 
so sorry, when i replied,i didn't see any answer in this thread.AngeloG 's reply will be ok.
 
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
 
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.
:smile:
 
  • #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
 
Back
Top