A point charge q_1 = 3.90 nC is placed at the origin, and a second point charge

Question

A point charge q_1 = 3.90 nC is placed at the origin, and a second point charge q_2 = -3.05 nC is placed on the x-axis at x =+ 21.0 cm. A third point charge q_3 = 2.10 nC is to be placed on the x-axis between q_1 and q_2. (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 q_3 is placed at x = + 11.0 cm?
U=______ J

(b) Where should q_3 be placed between q_1 and q_2 to make the potential energy of the system equal to zero?
x=_____ cm

Explanation / Answer

Given q 1 = 3.9 x 10 -9 C          q 2 = -3.05 x 10 -9 C          q 3 = 2.1 x 10 -9 C (a). The potential energy of the system of the three charges if q 3 is placed at x = + 11.0 cm is            U = [ -Kq 1 q 2 / 0.21m ] + [ -Kq 3 q 2 /0.1 m ] + [-Kq 1 q 3 / 0.11 m ] Since potential enenrgy due to two charges = -Kqq' / r Where K = Coulomb's cconstant                 = 8.99 x 10 9 Nm 2/ C 2            r = Separation between the charges Substitue values we get U = [5.092 x 10 -7 J ] +[ 5.758 x 10 -7 J ] + [ - 6.693 x 10 -7 J ]                                          = 4.157 x 10 -7 J (b). Potential energy is zero. Let at a distance x from the origin the potential energy is zero. U = [ -Kq 1 q 2 / 0.21] + [ -Kq 3 q 2 /(0.21-x) ] + [-Kq 1 q 3 / x ] 0 = [ -q 1 q 2 / 0.21] + [ -q 3 q 2 /(0.21-x) ] + [-q 1 q 3 / x ] 0 = [56.64 x 10 -18 ] + [ (6.405 x 10 -18 ) /(0.21 -x)] +[(-8.19  x 10 -18 ) / x ] 0 = 56.64 + [6.405 /(0.21-x) ] +[-8.19 / x] (8.19 / x) -[6.405 /(0.21-x) ]= 56.64 [(0.21-x) 8.19 -6.405 x ] / [x(0.21-x) ] = 56.64 [1.7199 -8.19 x -6.405 x ] /[ 0.21 x -x 2 ] = 56.64 1.7199 - 14.595x = 56.64 (0.21 x) - 56.64 x 2 56.64 x 2 - 26.4894 x +1.7199 = 0 Solve it   x = {26.4984 ± [(-26.4894) 2 -(4 x 56.64 x 1.7199 ) ] } / 2( 56.64 )                  = { 26.4984 ± 17.66} / 2 (56.64)                  = 0.3898 m    or 0.078 m So, answer is 0.078 m or 7.8 cm Since 0.3898 is not between two charges       (b). Potential energy is zero. Let at a distance x from the origin the potential energy is zero. U = [ -Kq 1 q 2 / 0.21] + [ -Kq 3 q 2 /(0.21-x) ] + [-Kq 1 q 3 / x ] 0 = [ -q 1 q 2 / 0.21] + [ -q 3 q 2 /(0.21-x) ] + [-q 1 q 3 / x ] 0 = [56.64 x 10 -18 ] + [ (6.405 x 10 -18 ) /(0.21 -x)] +[(-8.19  x 10 -18 ) / x ] 0 = 56.64 + [6.405 /(0.21-x) ] +[-8.19 / x] (8.19 / x) -[6.405 /(0.21-x) ]= 56.64 [(0.21-x) 8.19 -6.405 x ] / [x(0.21-x) ] = 56.64 [1.7199 -8.19 x -6.405 x ] /[ 0.21 x -x 2 ] = 56.64 1.7199 - 14.595x = 56.64 (0.21 x) - 56.64 x 2 56.64 x 2 - 26.4894 x +1.7199 = 0 Solve it   x = {26.4984 ± [(-26.4894) 2 -(4 x 56.64 x 1.7199 ) ] } / 2( 56.64 )                  = { 26.4984 ± 17.66} / 2 (56.64)                  = 0.3898 m    or 0.078 m So, answer is 0.078 m or 7.8 cm Since 0.3898 is not between two charges      

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