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JEB
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The potential energy of a uniform sphere of charge $Q$ is:

$$ U_Q = \frac 3 5 \frac 1 {4\pi\epsilon_0} \frac{Q^2}R $$

while the gravitational binding energy is:

$$ U_G = -\frac 3 5 \frac{GM^2} R $$

(Both of these results are standard physics problems).

You want to add them:

$$ U(R) = U_Q(R) + U_G(R) $$

and find the minimum with respect to $R$. Of course, they have the same behavior vs $R$, so that is pointless. One would think it is minimized at $R=\infty$.

However, if you fix $R$, it would be more interesting to the find $Q$ for which:

$$ U_Q = U_G$$

Note that in this treatment, the Sun is approximated as a gas of non interacting Newtonian particles, for which there are only two outcomes: explosion or implosion, depending on the ratio of $Q/M$.

You can use:

$$ -\frac{d(U_Q+U_M)}{dR}$$

to compute the inward or outward pressure.

Without putting in any numbers, if $q_p = 1.000001|e|$, the Sun would be obliterated in fashion that make gamma ray bursts seem dull, as the electric force is $10^{39}$ times stronger than gravity.

The potential energy of a uniform sphere of charge $Q$ is:

$$ U_Q = \frac 3 5 \frac 1 {4\pi\epsilon_0} \frac{Q^2}R $$

while the gravitational binding energy is:

$$ U_G = -\frac 3 5 \frac{GM^2} R $$

(Both of these results are standard physics problems).

You want to add them:

$$ U(R) = U_Q(R) + U_G(R) $$

and find the minimum with respect to $R$. Of course, they have the same behavior vs $R$, so that is pointless. One would think it is minimized at $R=\infty$.

However, if you fix $R$, it would be more interesting to the find $Q$ for which:

$$ U_Q = U_G$$

The potential energy of a uniform sphere of charge $Q$ is:

$$ U_Q = \frac 3 5 \frac 1 {4\pi\epsilon_0} \frac{Q^2}R $$

while the gravitational binding energy is:

$$ U_G = -\frac 3 5 \frac{GM^2} R $$

(Both of these results are standard physics problems).

You want to add them:

$$ U(R) = U_Q(R) + U_G(R) $$

and find the minimum with respect to $R$. Of course, they have the same behavior vs $R$, so that is pointless. One would think it is minimized at $R=\infty$.

However, if you fix $R$, it would be more interesting to the find $Q$ for which:

$$ U_Q = U_G$$

Note that in this treatment, the Sun is approximated as a gas of non interacting Newtonian particles, for which there are only two outcomes: explosion or implosion, depending on the ratio of $Q/M$.

You can use:

$$ -\frac{d(U_Q+U_M)}{dR}$$

to compute the inward or outward pressure.

Without putting in any numbers, if $q_p = 1.000001|e|$, the Sun would be obliterated in fashion that make gamma ray bursts seem dull, as the electric force is $10^{39}$ times stronger than gravity.

Source Link
JEB
  • 39.5k
  • 3
  • 42
  • 91

The potential energy of a uniform sphere of charge $Q$ is:

$$ U_Q = \frac 3 5 \frac 1 {4\pi\epsilon_0} \frac{Q^2}R $$

while the gravitational binding energy is:

$$ U_G = -\frac 3 5 \frac{GM^2} R $$

(Both of these results are standard physics problems).

You want to add them:

$$ U(R) = U_Q(R) + U_G(R) $$

and find the minimum with respect to $R$. Of course, they have the same behavior vs $R$, so that is pointless. One would think it is minimized at $R=\infty$.

However, if you fix $R$, it would be more interesting to the find $Q$ for which:

$$ U_Q = U_G$$