What is the work done to move this Charge?

[Reneee]

Homework Statement

A charge Q in point O=(0,0) and a test charge q in point A=(4,1) lie in a plane.
How much work W needs to be done to move q to point B=(2,2).
Q always stays at (0,0).
Q=q= 2*10^-4. One unit here equals 1 meter.

Relevant Equations

no equations.

I found two formulas to calculate the work done. One is with this path integral:
## W_{AB}## = W(## r_A,r_B ##)=q* ## \int_{r_A}^{r_B} E*dr ##
but here is the one I tried to use:
## W_{AB}## = q*Δ U = q*(## \frac {kQ} {r_A} ## - ## \frac {kQ} {r_B} ## )
Now here's my problem, what are the distances## r_A ##and ##r_B ## that i have to plug in?
Can I just use: the length ##r_A ##=## | \vec {OA} |## = ## \sqrt 17## ## r_B ## = ##| \vec {OB} | ## = ##\sqrt 8 ##.

Or do I have to do something else, like do I maybe need to follow the equipotential lines(please say no x) ) like this?:

If so ,then how can I put this into mathematics, especially the path ##\vec {BC}##, so I can use it in one of the equations?

~note: for all intents and purposes I didn't knew anything about physics until last week ,when we covered physics for 2 lessons in my cs-course, please excuse me if i am missing some of the very basics.

 

[TSny]

I found two formulas to calculate the work done. One is with this path integral:
## W_{AB}## = W(## r_A,r_B ##)=q* ## \int_{r_A}^{r_B} E*dr ##
but here is the one I tried to use:
## W_{AB}## = q*Δ U = q*(## \frac {kQ} {r_A} ## - ## \frac {kQ} {r_B} ## )

The second formula comes from the first formula after evaluating the integral. You can just use the second formula. However, the formula gives the work done by the electric force acting on q as the charge is moved from A to B. You are asked to find the work that "needs to be done" to move the charge from A to B. That is, how much work is done by your force if you move the charge from A to B? Can you see how the works done by these two different forces are related to one another?

Now here's my problem, what are the distances## r_A ##and ##r_B ## that i have to plug in?
Can I just use: the length ##r_A ##=## | \vec {OA} |## = ## \sqrt 17## ## r_B ## = ##| \vec {OB} | ## = ##\sqrt 8 ##.

Yes, just use these values for ##r_A## and ##r_B##. These are the initial and final distances between the two charges. The work is independent of the particular path, so you don't need to pick a path. (That's an important thing to keep in mind.) The work depends only on the initial and final positions, as indicated by the formula for ##W_{AB}## where you only need to know the initial and final ##r##'s.

 

[gneill]

I found two formulas to calculate the work done. One is with this path integral:
## W_{AB}## = W(## r_A,r_B ##)=q* ## \int_{r_A}^{r_B} E*dr ##
but here is the one I tried to use:
## W_{AB}## = q*Δ U = q*(## \frac {kQ} {r_A} ## - ## \frac {kQ} {r_B} ## )
Now here's my problem, what are the distances## r_A ##and ##r_B ## that i have to plug in?
Can I just use: the length ##r_A ##=## | \vec {OA} |## = ## \sqrt 17## ## r_B ## = ##| \vec {OB} | ## = ##\sqrt 8 ##.

Yes you can!

Or do I have to do something else, like do I maybe need to follow the equipotential lines(please say no x) ) like this?:
View attachment 255607
If so ,then how can I put this into mathematics, especially the path ##\vec {BC}##, so I can use it in one of the equations?

Note that moving a charge along an equipotential line requires no work. So the path CB requires no work to be done. You just have to get the particle from A to C, which just happens to be the change in radial distance (from Q) from A to that equipotential surface. So that's just moving from ##r_A## to ##r_B##.

Make sure that you get the correct sign for the work. Since both charges have the same sign, moving q closer to Q will require positive work.

~note: for all intents and purposes I didn't knew anything about physics until last week ,when we covered physics for 2 lessons in my cs-course, please excuse me if i am missing some of the very basics.

No worries there. We're here to help you. While we won't teach you an entire introductory physics course, we can help to point out misconceptions, spot errors, fill in the gaps in your understanding and so forth. Good luck!

Edit: Ha! TSny beat me to it!

 

[Reneee]

ah thanks guys I did it with point c and just like you said I got the same solution ,√8.