# electric field

Electrons 'leak' from the surface of many stars resulting in such stars acquiring a positive charge. This charging stops when the charge on the star is so large that protons in the surface also begin to be repelled. This occurs when the sum of the gravitational potential energy and the electric potential energy of a proton near the surface is zero.

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• 天同
Lv 7
9 年前
最愛解答

Electrons 'leak' from the surface of many stars resulting in such stars acquiring a positive charge.

Suppose at first the star is neutral. The escape of electrons from the star results in an unbalance of +ve and -ve charges. As the electrons (-ve charge) go away, leaving the protons (+ve charge) behind. Hence, the star is +ve charged.

This charging stops when the charge on the star is so large that protons in the surface also begin to be repelled.

Because like charges repel, proton and proton thus repel one anther. When the star becomes more and more +ve charged, the force of repulsion between protons increases.

This occurs when the sum of the gravitational potential energy and the electric potential energy of a proton near the surface is zero.

Protons are held together by gravitaional attractive force. The protons are stable when the gravtiational force is greater than the electrostatic repulsive force. As the force of electrostatic repulsion gradually increases, a point is reached when the two forces (gravitation and electrostatic) are equal.

That is, mathematically,

GMm/R^2 = kqQ/R^2

where G is the Universal Gravitatinal Constant

M is the mass of the star

m is the mass of a proton

R is the radius of the star

k is the electrostatic constant

q is the charge of a proton

Q is the total charge of the star

Hence, GMm/R = kqQ/R

i.e. kqQ/R - GMm/R = 0

kqQ/R + (-GMm/R) = 0

That means, the sum of electrostatic potential (the first term) of a proton and its gravitational potential (the second term, remember that gravitational potential is -ve due to its attrative nature) is zero.