# Electromagnetic black hole?

So I was thinking about something for the past while

Consider a large spherical foam-ball with homogeneous density. Where a foam ball is defined as an object that can absorb matter with 0 friction (example: a gravitational well without an object inside). This is a purely theoretical construct.

If the foam-ball has a radius R and a charge Q. What charge must the foam-ball have, such that there is well defined sphere or horizon such that any object with a negative charge (even if it is the slightest bit) must travel faster than C in order to escape the field of the ball.

Aka what charge would turn this into a black-hole for all objects with opposite sign on their charge?

Once again I assume the foam-ball is a single particle, that doesn't repel itself... its just a very large homogenous 'thing' with charge.

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Can this approached using a similar style of math as the construction of a gravitational black hole, except mass is substituted with charge, etc... ? – frogeyedpeas May 28 '13 at 19:14
I think you should just use basic equations as $F_e = k*\frac{Q_1*Q_2}{r^2}$ and $a = \frac{F}{m}$. You then put e constant instead of $Q_2$. I'm only not sure how to deal with aceleration. – Tomáš Zato May 28 '13 at 19:24

## 1 Answer

Such a singularity would not occur, if you have no lower bound on the negative charge. For gravity, the singularity occurs because gravitational potential energy and relativistic kinetic energy both depend on the mass of the smaller object, which allows it to divide out when you solve for escape velocity. However, in this case, only the electromagnetic potential energy, not the kinetic energy, depends on the negative charge. This means the negative charge never gets divided out, and hence you can arbitrarily decrease the magnitude of this charge to make the escape velocity as low as you want.

However, if you fix the negative charge at some value $q_2$, it will be unable to escape when its electromagnetic potential energy equals its rest energy (up to a sign). So just set $\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r} = mc^2$ (again, up to the correct sign) and solve for $r$.

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do u think u might have a link or something that explains how the gravitational potential energy and relativistic kinetic energy 'divide out'? I'm curious if there is an analog of kinetic energy that applies to E&M. – frogeyedpeas May 28 '13 at 19:44
Yep, try here. It explains how to derive relativistic escape velocity in a gravitational field, and you should see $m_0$ divide out in the algebra. – Izzhov May 28 '13 at 19:46
Looks good! this is much clearer than I expected. – frogeyedpeas May 28 '13 at 19:55