A point charge q1 = 6.00 nC is placed at the origin, and a second point charge q

Question

A point charge q1 = 6.00 nC is placed at the origin, and a second point charge q2 = ?4.50 nC is placed on the x-axis at x = +30.40 cm. A third point charge q3 = 3.00 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.) (a) What is the potential energy of the system of the three charges if q3 is placed at x = +15.20 cm? J (b) Where should q3 be placed to make the potential energy of the system equal to zero? x = m

Explanation / Answer

Please change the values 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 ______________________________________… 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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