A point charge q1 395 nC is placed at the origin and a seco

A point charge q_1 = 3.95 nC is placed at the origin, and a second point charge q_2 = -2.90 nC is placed on the x-axis at x =+ 20.0 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

Solution

a) Potential energy = potential enery between q1 and q2+ potential energy between q2 and q3 +potential energy between q1 and q3

= kq1q2/r1+kq1q3/r2+kq2q3/r3 = k(-3.95*2.90/20 + 3.95*2.10/11 -2.90*2.10/9) *10^-9/10^-2 =-4.46 * 10^-2 J

b) kq1q2/r1+kq1q3/r2+kq2q3/r3 =0

let distance be x from q1

so -3.95*2.90/20 + 3.95*2.10/x -2.90*2.10/(20-x)=0

solving for x we get x=7.75 cm.