12n Problem 8: Three point charges Qi, Q2, and Qs are placed on the corners of a

Question

12n Problem 8: Three point charges Qi, Q2, and Qs are placed on the corners of an equilateral triangle and Qi, Q2, are held fixed (see diagram). 2 Assume that Qi and Q2 are protons and Qs is an electron, and L = 1.25 mm. L/2 a) What is the electric potential at point A? b) What is the electric potential energy of the electron Qs at the L/2 Q2 position shown in the diagram? velocity at point A? answer either by a calculation or in one sentence. c) if Qs- the electron - is released from rest, what will be its d) Where will the electron momentarily stop? Justify your 3

Explanation / Answer

8)Givevn,

Q1 = Q2 = 1.6 x 10^-19 ; Q3 = -1.6 x 10^-19 ; L = 1.25 mm = 1.25 x 10^-3 m

a)The potential at a will be the algebric sum of the individual potentials due to the charges.

Va = kQ1/r1 + kQ2/r2 + kQ3/r3

r1 = r2 = 1.25/2 = 0.625 x 10^-3 m ; r3 = sqrt (1.25^2 - 0.625^2) = 1.08 x 10^-3 m

Va = 9 x 10^9 x 1.6 x 10^-19 (2/0.625 x 10^-3 - 1/1.08 x 10^-3) = 3.27 x 10^-6 V

Hence, Va = 3.27 x 10^-6 V

b)We first need to know the potential at the location of Q3

V3 = k Q1/r1 + kQ2/r2 = 2 Q1/r1

V3 = 2 x 9 x 10^9 x 1.6 x 10^-19/(1.25 x 10^-3) = 2.304 x 10^-6 V

PE = Q3 V3

PE = -1.6 x 10^-19 x 2.304 x 10^-6 = -3.69 x 10^-25 J

Hence, PE = -3.69 x 10^-25 J

c)from conservation of energy

KE = PE

1/2 m v^2 = 3.69 x 10^-25

v = sqrt (2 x 3.69 x 10^-25/9.1 x 10^-31) = 901 m/s

Hence, v = 901 m/s

d)The electron will mometarily stop at point A. Since it will experience attractive forces from the two protons placed at either sides of the pint A.

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