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Work done to move a test charge

  • Thread starter ND3G
  • Start date
  • #1
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A test charge of +1.0*10^-6 C is 40cm from a charged sphere of 3.2*10^-3

How much work is required to move it there from a point 1.0*10^2 m away from the sphere?

GIVEN:
q1 = +1.0*10^-6 C
q2 = 3.2*10^-3 C
r1 = 1.0*10^2 m
r2 = 0.40 m

Required:
W

Analysis:
Ee=(k*q1*q2)/r
delta Ee=E2-E1
W=delta Ee

Solution:

179.99712 J

Am I on the right track here? It just doesn't seam right to me for some reason. Also, since the charged sphere does not clearly specify that it is a positive or negative charge, should I assume it is postive?
 

Answers and Replies

  • #2
G01
Homework Helper
Gold Member
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16
I am assuming by Ee, you mean Electrical potential energy, correct?

If this is the case then I believe you are on the right track.


G01
 
Last edited:
  • #3
66
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Can someone help me with this question. please. i have tried all i could but i am nowhere arround the right answer. :(
 
  • #4
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I don't know where the 179.99712 J came from. By my calculation k*q1*q2 is (9.0x10^9)*(1.0*10^-6)*(3.2*10^-3)=28.8 According to this Ee2 = 28.8/0.4=72 and Ee1 = 28.8/100 = 0.288
 
  • #5
Defennder
Homework Helper
2,591
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Can someone help me with this question. please. i have tried all i could but i am nowhere arround the right answer. :(
I got the answer as 179.75 J, which is pretty close to the OP's answer. You should post your attempt so we can check it for you.
 
  • #6
37
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I don't think the OP's answer is correct. Defennder do you mind sharing how you got that number?

Here's what I got

GIVEN:
q1 = +1.0*10^-6 C
q2 = 3.2*10^-3 C
r1 = 1.0*10^2 m
r2 = 0.40 m

E1 = (k*q1*q2) / r1
= (9.0*10^9*3.2*10^-3*1*10^-6)/(100m)
= 0.288 J

E2 = (k*q1*q2) / r2
= (9.0*10^9*3.2*10^-3*1*10^-6)/(0.4m)
= 72 J

W=deltaE= E2 - E1
= 72J - 0.288J
= 71.712J

Therefore 71.712J of work is required.

Can anyone comment on this please? I honestly don't get how they got a number close to 180J
 
  • #7
Doc Al
Mentor
44,892
1,144
Therefore 71.712J of work is required.

Can anyone comment on this please?
Looks good to me.
 
  • #8
Defennder
Homework Helper
2,591
5
Well this thread has been inactive for some time. Looking back I think my error and probably that of the OP was to square 0.4m Otherwise I think the correct answer should be about -71.6J.
 
  • #9
11
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For this very same question above, how would I determine the electrons that were gained or lost from the test subject to create the charge?
 
  • #10
Doc Al
Mentor
44,892
1,144
For this very same question above, how would I determine the electrons that were gained or lost from the test subject to create the charge?
Assuming the test object was originally neutral, every time you remove one electron the charge remaining will increase by one elementary charge (which is the magnitude of the charge on an electron). What's the charge on an electron?
 
  • #11
11
0
1.60217646 × 10-19
 
  • #12
11
0
do I multiply the test charge with 6.242 × 10^18 e using millikan's relationship?
 
  • #13
Doc Al
Mentor
44,892
1,144
do I multiply the test charge with 6.242 × 10^18 e using millikan's relationship?
That works. Realize that multiplying by that number is the same as dividing by the electron charge.
 
  • #14
11
0
oh okay! simple enough. thanks a bunch.
 

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