# what is Q in E=kQ/r^(2)

ik that in V= W/Q, Q is the charge that is being moved in an electric field.

But in E=kQ/r^(2), Q is the charge that creates an electric field??
so in the case where we direct a high energy alpha particle to a gold nucleus, whose Q are we going take in the equation: E= kQ/r^(2) ?

aren't the electric fields of both the aplha particle and gold nucleus interacting and producing a final electric field? so why do we then, only take charge of the gold in the equation?
(edited 9 months ago)
Original post by Aleksander Krol
ik that in V= W/Q, Q is the charge that is being moved in an electric field.

But in E=kQ/r^(2), Q is the charge that creates an electric field??
so in the case where we direct a high energy alpha particle to a gold nucleus, whose Q are we going take in the equation: E= kQ/r^(2) ?

aren't the electric fields of both the aplha particle and gold nucleus interacting and producing a final electric field? so why do we then, only take charge of the gold in the equation?

I think you'd have to take the Q of the gold nucleus because essentially the alpha particle is entering the field produced by the gold nucleus. I also think that because the gold nucleus is heavier, the magnitude of it's charge will be bigger meaning the field strength it produces will be larger as well which is why you only take the charge of the gold. That's the only explanation that makes sense to me.
Original post by Aleksander Krol
ik that in V= W/Q, Q is the charge that is being moved in an electric field.

But in E=kQ/r^(2), Q is the charge that creates an electric field??
so in the case where we direct a high energy alpha particle to a gold nucleus, whose Q are we going take in the equation: E= kQ/r^(2) ?

aren't the electric fields of both the aplha particle and gold nucleus interacting and producing a final electric field? so why do we then, only take charge of the gold in the equation?

But in E=kQ/r^(2), Q is the charge that creates an electric field??

Yes, indeed the Q in the electric field formula is the charge that creates the electric field around the charge.

aren't the electric fields of both the alpha particle and gold nucleus interacting and producing a final electric field? so why do we then, only take charge of the gold in the equation?

Yes, both the charged alpha particle and gold nucleus would produce a resultant field at points around them.
That resultant electric field at that point is meant for calculating the resultant interaction between gold nucleus and another charged particle at that point and between charged alpha particle and another charged particle at that point.

$F = \dfrac{kQ_{\alpha}q}{r_1^2} + \dfrac{kQ_{Au}q}{r_2^2}$
where
r1 is the separation between the alpha particle and the other charged particle,
r2 is the separation between the gold nucleus and the other charged particle,
Qα is the charge of alpha particle,
QAu is the charge of gold nucleus.

Key idea is the alpha-charged particle interacts via the electric field of the gold nucleus.