Electric potential of point charge

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pyninja
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Homework Statement


A charged particle is fixed in place at the origin. A second particle of charge [tex]+10^{-6}[/tex] C is released from rest from very far away ([tex]\approx (\infty, 0)[/tex]). The second particle passes the point (9 m, 0) with a kinetic energy of 1.0 J.

Find the electric potential due to the fixed charged particle at the point (9 m, 0).

Homework Equations


[tex]F = \frac{q_1q_2}{r^2} = ma[/tex]
[tex]V = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r}[/tex]


The Attempt at a Solution


We need to the find the charge of the fixed particle to apply the formula for V... I tried using conservation of energy, but it seems to me that the initial potential energy and kinetic energy are both 0. Not sure what I'm doing wrong there...
 
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ok why don't you use foruma U= Q*delta V if as you say the total energy is zero then U2 = -K2
 
madah12 said:
ok why don't you use foruma U= Q*delta V if as you say the total energy is zero then U2 = -K2

Oh, for some reason I thought the energy has to be positive, haha.

So then I get [tex]q_1 = -\frac{k r}{q_2}[/tex].

And [tex]V = -\frac{k^2}{q_2}[/tex]?

(where [tex]k = \frac{1}{4\pi\epsilon_0}[/tex])

Thanks for your help!
 
Only kinetic energy has to be positive.