A point charge q1= 3.90nC is placed at the origin, and a second point charge q2 = -3.00nC is placed on the x-axis at x=+20.0 cm. A third point charge q3= 2.00nC is to be placed on the x-axis between q1 and q2. (Take as zero the potential energy of the three charges when they are infinitely far apart.)

**Answer 1**

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A point charge q1= 3.90nC is placed at the origin,

Second point charge q2 = -3.00nC is placed on the x-axis at x=+20.0 cm.

Third point charge q3= 2.00nC is placed at x=+ 10.0 cm

Potential energy = 9*10^ 9[ -58.5*10^-18 – 60.0 *10^-18 +78 *10^-18 ]

Potential energy = 9*10^- 9[ -40.5 ]

Potential energy = 3.645*10^- 7 J

A) The potential energy of the system of the three charges if q3 is placed at x=+ 10.0 cm is 3.645*10^- 7 J

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For potential energy to be zero, let q3 be placed at ‘x’

Potential energy = 9*10^ 9[ -58.5*10^-18 – 6.0/ (0.2 – x) *10^-18 +7.8/x *10^-18 ]

Potential energy = 9*10^ -9[ -58.5 – 6.0/ (0.2 – x) +7.8/x ]= 0

[-58.5 – 6.0/ (0.2 – x) +7.8/x ]= 0

[ – 6.0/ (0.2 – x) +7.8/x ]= 58.5

[ – 6.0*x +7.8(0.2 – x) ]= 58.5(0.2 – x)x

[ – 6.0*x +1.56 -7.8 x ]= 11.64x -58.5x^2

58.5x^2 – 25.44 x +1.56 = 0

x = [25.44 (+ -) sq rt (647.1936-365.04 )]/117

x = [25.44 (+ -) sq rt 282.1536] / 117

x = [25.44 (+ -) 16.797 ] /117

x =0.35 m or x =0.074 m

As q3 is to be placed in between q1 and q2, we select x = 0.074 m or x =7.4cm

B) q3 should be placed between q1 and q2 at x =7.4 cm from origin to make the potential energy of the system equal to zero

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