A point charge q1 350 nC is placed at the origin and a seco

A point charge q_1 = 3.50 nC is placed at the origin, and a second point charge q_2 = -2.0 nC is placed on the x-axis at x = +15.00 cm. A third point charge q_3 = 1.75 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). What is the potential energy of the system of the three charges if q_3 is placed at x = + 7.50 cm Where should q_3 be placed to make the potential energy of the system equal to zero.

Solution

given that,

q1 = 3.50 nC, at x1 = 0;

q2 = -2.00 nC, at x2 = 15 cm = 0.15 m,

q3 = 1.75 nC, at x3 = 7.5 cm = 0.075 m

the potential energy U of the system is,

U = kq1*q2/|x2 - x1| + kq1*q3/|x3 - x1| + kq2*q3/|x2 - x3|

   = -1.05e-7 J

b) use the below condition and solve the value of x3.

    let, U = 0

      U = kq1*q2/|x2 - x1| + kq1*q3/|x3 - x1| + kq2*q3/|x2 - x3| = 0

then,

         q1*q2/|x2 - x1| + q1*q3/|x3 - x1| + q2*q3/|x2 - x3| = 0

    3.5*(-2)/0.15 + 3.5*1.75/|x3| + (-2)*1.75/|0.15 - x3| = 0

                         -46.67 + 6.125/|x3| - 3.50/|0.15 - x3| = 0

if x3 < 0, -46.67 + 6.125/|x3| - 3.50/|0.15 - x3| = 0,

       values of three terms are positive, so there is no solution.

if 0 < x3 < 0.15 m, -46.67 + 6.125/|x3| - 3.50/|0.15 - x3| = 0, then x3 = 0.0684 m = 6.84 cm

if x3 > 0.210 , -46.67 + 6.125/|x3| - 3.50/|0.15 - x3| = 0,, no solution.

therefore, the required position is, x3 = 0.0684 m