Electric potential vs gravitational potential and nuclear collision limits

[thomas49th]

I always try and think about the gravatational laws when thinking of the electrical ones:
Can I just confirm that

Electrical field strength in a uniform field = F/q (force per unit of charge)

A electrical field strength at distance r from the body is $$\frac{-kQ}{r^{2}}$$
where Q is the charge of the body and k = $$\frac{1}{4\pi E_{0}}$$

Electrical potential (I've been looking on wiki but what actually does this potential do) = -kM/r, where M is the of body that gives out the charge at a

F = $$\frac{-kQq}{r^{2}}$$

but like gravity is there a E = $$\frac{-kQq}{r}$$ ?

I'm having a bit of a problem with potential. In exam questoins they keep referring to the electric potential then just 'the potential', but that would mean gravity wouldn't it, by default don't we say potential is GPE. The mark schemes have no mention of gravity...is my common sense playing up?

I know about potential wells (and thus why we define the attractive force as -, to show a body is bound or under the effect of another body)

The question I have shows a 8MeV alpha particle headnig towards a gold-197 nucleus and being turned back (distance of closet approach type thing). How would I go about calculating the upper limit on the size of the gold nucleus? First I'm guessing I need to find the energy of the alpha particle at point closet to the nucleus?? I would expect the answer to be ~ $$10^{-14}$$


Thanks :)

 

[LowlyPion]

The question I have shows a 8MeV alpha particle headnig towards a gold-197 nucleus and being turned back (distance of closet approach type thing). How would I go about calculating the upper limit on the size of the gold nucleus? First I'm guessing I need to find the energy of the alpha particle at point closet to the nucleus?? I would expect the answer to be ~ $$10^{-14}$$


Thanks :)

To address your question, the energy in Mev is an expression of the initial kinetic energy of the particle right? And to figure where that energy - hence its velocity is depleted and then reversed - where the kinetic energy converts to electrical potential energy - you need to examine where the change in potential energy is such that the work that goes into building that potential energy (of repulsion in this case) is equal to the energy the particle originally had.

For electrical potential it is expressed as the work to bring a particle from infinity (where it is 0) to the point of interest (if it is repulsive) and in the case where it is attractive (gravity is attractive too and hence is treated similarly), then it carries a negative sign, because the force involved over the distance needs to be used to resist the particle from wanting to accelerate nearer.

 

[Redbelly98]

I'm having a bit of a problem with potential. In exam questoins they keep referring to the electric potential then just 'the potential', but that would mean gravity wouldn't it, by default don't we say potential is GPE. The mark schemes have no mention of gravity...is my common sense playing up?

If the problem is one involving electrical charge, "the potential" is indeed referring to electrical potential (units are Volts or J/C), and not gravitational potential energy.

 

[diazona]

but like gravity is there a E = $$\frac{-kQq}{r}$$ ?

Yep, it's potential energy. Gravitational potential energy is
$$U = -\frac{GMm}{r}$$
and electrical potential energy is
$$U = -\frac{kQq}{r}$$

But electrical potential (as opposed to potential energy) is
$$V = -\frac{kQ}{r}$$
which is potential energy per unit charge. (That is, potential energy of a "test charge" per unit test charge) This is the same quantity you probably know as "voltage."