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diazona is offline
Jun25-09, 03:32 PM
HW Helper
P: 2,156

As for your question about protons... if I understand correctly, you're asking how many protons it would take to produce an electric force equivalent to Earth's gravity? That depends on what's being attracted to the protons.

Here's how you'd figure it out: the formula for gravitational force is
[tex]F_G = G \frac{Mm}{r^2}[/tex]
and the one for electrostatic force is
[tex]F_E = k \frac{Qq}{r^2}[/tex]
In this situation, G is the universal gravitational constant, M is the mass of the Earth, k is the electrostatic constant, and Q is the charge of the "electric Earth equivalent" (which is what you're asking for).
[tex]\begin{align*}G &= 6.67\times 10^{-11} \frac{\mathrm{N}\ \mathrm{m}^2}{\mathrm{kg}^2} \\
M &= 5.97\times 10^{24}\ \mathrm{kg} \\
k &= 8.99\times 10^{9} \frac{\mathrm{N}\ \mathrm{m}^2}{\mathrm{C}^2}\end{align*}[/tex]
Now, if we're saying that the electrostatic force is equivalent to gravity,
[tex]F_E = F_G[/tex]
[tex]k \frac{Qq}{r^2} = G \frac{Mm}{r^2}[/tex]
[tex]Q = \frac{GM}{k}\frac{m}{q}[/tex]
Now, you can easily calculate [itex]\frac{GM}{k}[/itex] by just multiplying/dividing those numbers I mentioned. But [itex]\frac{m}{q}[/itex] depends on the particle that's being attracted to the Earth or ball of protons. For example, if it were an electron, you'd use the mass of the electron [itex]m = 9.1\times 10^{-31}\ \mathrm{kg}[/itex] and its charge [itex]q = -1.6\times 10^{-19}\ \mathrm{C}[/itex], do the calculations, and find that it would take [itex]2.5\times 10^{-7}\ \mathrm{C}[/itex] of charge to attract an electron as hard as the Earth does. That's just over a trillion protons, which is practically nothing - a millionth of a speck of dust, maybe.