A point charge q 1 = 3.90 nC is placed at the origin, and a second point charge

Question

A point charge q1 = 3.90 nC is placed at the origin, and a second point charge q2 = -3.05 nC is placed on the x-axis at x=+ 20.5 cm . A third point charge q3 = 2.10 nC 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.)

Part A

What is the potential energy of the system of the three charges if q3 is placed at x=+ 10.5 cm ?

A point charge q1 = 3.90 nC is placed at the origin, and a second point charge q2 = -3.05 nC is placed on the x-axis at x=+ 20.5 cm . A third point charge q3 = 2.10 nC 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.)

Part A

What is the potential energy of the system of the three charges if q3 is placed at x=+ 10.5 cm ?

Explanation / Answer

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

Second point charge q2 = -3.05nC is placed on the x-axis at x=+20.5 cm.

Third point charge q3= 2.10nC is placed at x=+ 10.5 cm

Potential energy = 9*10^ 9[ -53.86*10^-18 - 53.86 *10^-18 +72.43 *10^-18 ]

Potential energy = 9*10^- 9[ -35.29 ]

Potential energy = 3.1761*10^- 7 J

A) The potential energy of the system of the three charges if q3 is placed at x=+ 10.5 cm is 3.1761*10^- 7 J

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

Potential energy = 9*10^ 9[ -53.86*10^-18 - 5.655/ (0.21 - x) *10^-18 +7.605/x *10^-18 ]

Potential energy = 9*10^ -9[-53.86 -5.655 / (0.21 - x) +7.605 / x ]= 0

[-53.86 -5.655 / (0.21 - x) +7.605 / x ]= 0

[ -5.655/ (0.21 - x) +7.605 / x ]= 53.86

[ - 5.655*x +7.605 (0.21 - x) ]= 53.86(0.21 - x)x

[ -5.655*x +1.597 -7.605 x ]= 11.3106x -53.86x^2

53.86x^2 - 24.57 x +1.597 = 0

x = [24.57 (+ -) sq rt (603.71-344.06 )]/107.72

x = [24.57 (+ -) sq rt 259.65] / 107.72

x = [24.57 (+ -) 16.11 ] /107.72

x =0.38 m or x =0.0785 m

As q3 is to be placed in between q1 and q2, we select x = 0.0785 m or x =7.85cm

B) q3 should be placed between q1 and q2 at x =7.85 cm from origin to make the potential energy of the system equal to zero

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