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Work done to move a charge

  • Thread starter rojasharma
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66
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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??
 

Answers and Replies

Work is defined as F*ds. Look at the equations you used and the units they give.
 
66
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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 thats the right approch here:S)
 
171
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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.
 
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Don;t see that problem in cramster.com:(, thanks for the web though.
 
Doc Al
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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.
 
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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??)
 
Doc Al
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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.
 
66
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Oh, sorry the E1 is 72J.....so its delta E= E1- E2???/ how did i mix up?
 
Doc Al
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Initial position: r1 = 100 m.
Final position: r2 = 0.4 m.
 
66
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right ...silly me...thanks a lot
 
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.
 
Doc Al
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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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