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What is the number of excess electrons on each sphere in the following case?Two small...

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lexijuly | Student, Undergraduate | (Level 1) Honors

Posted December 28, 2010 at 4:58 PM via web

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What is the number of excess electrons on each sphere in the following case?

Two small spheres spaced 35 cm apart have an equal charge. How many excess electrons must be present on each sphere so that the magnitude of the force of repulsion between them is 2.20*10^21 N.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted December 28, 2010 at 5:05 PM (Answer #1)

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We'll write the Coulomb force formula:

F = (q^2/r^2)*K, where k is electrostatic constant; k = 8.988*10^9Nm^2C^-2.

We know the force F and the the distance between spheres:

F = 2.20*10^21N and r = 0.35m (conversion from cm to m)

F*r^2*K = q^2

q = sqrt (F*r^2*K)

q = sqrt (2.20*10^21*0.1225*8.988*10^9)

q = 0.17*10^-15 C

The charge of a single electron is e = 1.6*10^-19 C, so the number of excess electrons is:

N = q/e

N = 0.17*10^-15 C/1.6*10^-19 C

N = 1062 approx.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted December 28, 2010 at 5:27 PM (Answer #2)

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The force between two bodies with a charge q1 and q2 and a distance of separation r is given by Coulomb's law as q1*q2/(4*pi*e0*r^2). 1/(4*pi*e0) = 8.987*10^9 Nm^2/C^2

Now the distance between the spheres in the problem is 35 cm or 0.35 m.

The force of repulsion is supposed to be 2.20*10^21 N.

Now the charge on an electron is −1.602* 10^-19 C

If there are n electrons on each sphere:

n^2* (-1.602*10^-19)^2*8.987*10^9/0.35 = 2.20*10^21

=> n^2 = 2.2* 10^21*.35 / (-1.602*10^-19)^2*8.987*10^9

=> n^2 = 3.3*10^48

So n can be taken to be approximately 1.8*10^24.

Therefore each sphere should have an excess of 1.8*10^24 electrons.

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