Superposition principle and charges from infinity

[josef1234]

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.

 

[nickjer]

You need to divide by 'r' in your potential equation. Also, 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.<br /> <br /> EDIT: Fixed the above work equation.

 

[josef1234]

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.

 

[nickjer]

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:

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}

since r_i = \infty[/tex]

 

[josef1234]

Thanks nickjer,

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

Regards
Joe