Can someone please draw and explain the electric potential vs distance graphs for a positive and negative point charge? PLEASE. I really need help with this concept.

Question

anonymous · Answer

http://www.personal.psu.edu/faculty/b/q/bqw/physics_202/phy_202_3/phy202_3_pp/sld012.htm This webpage shows is +ve for both attractive and repulsive forces. But my book says it is +ve for repulsive force and -ve for attractive force. What is right?

anonymous · Answer

"This webpage shows is +ve for both attractive and repulsive forces. But my book says it is +ve for repulsive force and -ve for attractive force. What is right?"

this made no sense ot me


if you have a positive charge.. potential everywhere is positive.. decreases 1/r as you move away from it

and if you have a negative charge, potential everywhere is negative.. INCREASES 1/r as you move away from it (increases cause, it decreases in negative.. )

anonymous · Answer

Umm, I'm sorry. I missed the word potential. From what you have described, I think the webpage is wrong then. :O

anonymous · Answer

i have no clue.. wht the website is trying to convey xD

anonymous · Answer

The graph in the figure shows the electric potential energy as a function of separation for two point charges.
If one charge is +0.44 {\rm nC}, what is the other charge?

anonymous · Answer

It's okay. Can you help me with this?

anonymous · Answer

That's +0.44 nC.

anonymous · Answer

Should I simply use the formula? 
Like, $$U=\frac{ 1 }{ 4 \pi \epsilon } \frac{ q (0.44 \times 10^{-9}) }{ (4 \times 10^{-2})^{2} }$$

anonymous · Answer

I used the point $$(-1 \mu J, 4cm)$$

anonymous · Answer

@Mashy : Is this the right way to do it?

anonymous · Answer

wat lemme see the question

anonymous · Answer

its NOT square

and since the potential energy is negative.. the other charge MUST be +ve.. 

potential energy formula is

U = kQ1Q2/r

anonymous · Answer

Oh, right. Thank you so much.

anonymous · Answer

Should't the other charge be negative?

anonymous · Answer

oh yea.. sorry.. i thought the first charge is negative

the product should be negative.. so you are right.. my bad :P

anonymous · Answer

Thank you once again. ^_^

anonymous · Answer

no problem!