Superposition principle and charges from infinity

  • Thread starter josef1234
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  • #1
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This is not a homework assignment but a question from a past exam paper as i am studying for an exam on monday. I am one of only 3 out of 150 who have never done physics before this, the first year of an electronics degree. I would really appreciate any help.

1) A charge q1=1e-4C is in a fixed location at the origin r1=0. Calculate the energy required to transport a charge q2=2e-5C from infinity to the location r2=(0.1i + 0.2j)m

2) With q1 and q2 at the fixed locations r1 and r2, calculate the additional energy required to transport a third charge q3=1e-5C from infinity to the location r3=(0.2i + 0.1j)



V=(ke*Q)r

U=q(V1-V2)


I am very confused as to which values for Q i use in the equations and also what is the value for r if the point is at the origin, in fact having tried for almost three hours i am no closer to attaining an understanding.
 

Answers and Replies

  • #2
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You need to divide by 'r' in your potential equation. Also, [itex]W=\Delta V[/tex] is a useful equation for both problems. Just compute the electric potential on your test charge from the fixed charges. And then compute the change in potential energy from its two positions.

EDIT: Fixed the above work equation.
 
Last edited:
  • #3
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Thanks for your reply,

That was a typo in the equation - thanks for pointing it out!

So i have V=(keQ)/r but what is my r! i.e in the equation there is a q and an r but r relates the distance between two point charges so which value for q do you use?

Following further study i have attempted the question in a different way:

W=integral(F'.ds') F'=((ke*Q*q)/r^2)r' so W=(ke*q1*q2)/r

q1=1e-4C q2=2e-5C r=sqrt(0.1^2+0.2^2)=0.224

therefore W=3.579E-16 J

How'd i do? even if i am right could someone explain my quiry above, thank you.

For the second part of the question do i just use the above equation for W(q1q3) and W(q2q3) and add them together, is this what the superposition principle tells us.
 
  • #4
674
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That looks correct. And it is the same solution you would have got if you just used U = q*V. Since in your case q = q2 and V = k*q1/r, leaving you with V = k*q1*q2/r.

To calculate the work done, you would use:

[tex]W = \Delta V = kq_1 q_2 \left(\frac{1}{r_f}-\frac{1}{r_i}\right) = kq_1 q_2\left(\frac{1}{r_f} - 0\right) = \frac{kq_1 q_2}{r_f}[/tex]

since [itex]r_i = \infty[/tex]
 
  • #5
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Thanks nickjer,

That makes sense now, at last! here's hoping it comes up on monday....

Regards
Joe
 

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