How Is Work Calculated in Electric Fields?

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Homework Help Overview

The discussion revolves around calculating work done in electric fields, specifically involving a test charge and a charged sphere. Participants are exploring the relationship between electric potential energy and work in the context of electrostatics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of equations for electric potential energy and the calculation of work done. There are attempts to clarify the correct approach and units involved. Some participants question the initial and final positions of the charges and the implications of negative values in their calculations.

Discussion Status

The discussion is ongoing, with participants providing insights and corrections to each other's calculations. There is a mix of attempts to clarify concepts and explore different methods, such as integrating force over distance to derive expressions for electric potential energy.

Contextual Notes

Participants reference a similar problem from a textbook and express confusion over discrepancies in their results. There is mention of external resources like cramster.com, which some participants find unhelpful. The problem's parameters, including distances and charge values, are being scrutinized for accuracy.

rojasharma
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a test charge of +1.0x10^-6C is 40cm from a charged sphere of 3.2x10^-3C. A) how much work was required to move it therer from a point 1.0x10^2m away from the sphere? b) how many electrons were gained or lost from the test object to creat the charge?. I tried but i do not get the right answer:(. what i did was..used Ee1=kq1q2/r1 (r1 as 0.4m), Ee2=kq1q2/r2(r2 as 100m), then found delta Ee= Ee2=Ee1, Delta Ee= W??
 
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Work is defined as F*ds. Look at the equations you used and the units they give.
 
In the book there is excatly same question except it is moved 100cm instead of 100m. And the answer they got was 43 J. i tired all possible ways...i don;t get the right answer. if i use the above equation, delta Ee=E2-E1, it gives me 43..( but i don;t think that's the right approch here:S)
 
I think you need another equation to solve for this. First, try to find that problem on cramster.com (they have solutions for many problems). We are also learning Electric Fields(physics 11) in my class at the moment.
 
Don;t see that problem in cramster.com:(, thanks for the web though.
 
rojasharma said:
what i did was..used Ee1=kq1q2/r1 (r1 as 0.4m), Ee2=kq1q2/r2(r2 as 100m), then found delta Ee= Ee2=Ee1, Delta Ee= W??
That sounds right. You need to find the change in electric potential energy as the charges are brought together.
 
So,
q1=3.2x10^-3C
q2=1x10^-6C
r1=0.4m
r2=100m

Solution
E1= 72J (E1= (kq1q2)/r1)
E2= 0.288J (E2=kq1q2/r2)

delta E= E2-E1
delta E= 0.288J - 180J
delta E=-179.712J
Delta E= work done...(what about the negative value??)
 
rojasharma said:
So,
q1=3.2x10^-3C
q2=1x10^-6C
r1=0.4m
r2=100m

Solution
E1= 72J (E1= (kq1q2)/r1)
E2= 0.288J (E2=kq1q2/r2)
Looks OK. (You seemed to have switched r1 & r2: the charge moves from 100m to 0.4m.)

delta E= E2-E1
delta E= 0.288J - 180J
delta E=-179.712J
Delta E= work done...(what about the negative value??)
Where did you get 180J from? The negative sign is due to you mixing up initial and final positions.
 
Oh, sorry the E1 is 72J...so its delta E= E1- E2?/ how did i mix up?
 
  • #10
Initial position: r1 = 100 m.
Final position: r2 = 0.4 m.
 
  • #11
right ...silly me...thanks a lot
 
  • #12
Doc Al said:
That sounds right. You need to find the change in electric potential energy as the charges are brought together.

hmm. I am thinking of writing the force between the two charges and integrating it over the distance.
 
  • #13
PaintballerCA said:
I am thinking of writing the force between the two charges and integrating it over the distance.
Which is how one would derive the expression for electric potential energy.
 

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