Question

Answer the question which is in attachment ​

Answer 1

Explanation:

Answer: We can drop the "micro-coulomb" without any change in the problem. So we have, using Coulomb's law:

2/r^2 +-6/(15-r)^2 = 0, valid only for 0 < r< 15

2/r^2 = 6/(15-r)^2

Vol 86

2*(15-r)^2 = 6*r^2

2*(r^2-30r+225) = 6*r^2

r^2-30r+225 = 3*r^2

2r^2+30r-225 = 0

Vol 86

Apply the quadratic formula:

r = (-30 +/- sqrt(30^2 + 4*2*225))/(2*2)

r = (-30 +/- sqrt(2700))/4

= (-30 +/- 30*sqrt(3))/4

r = (15/2)*(-1 +/- sqrt(3)), valid only for 0 <r<15

r = (15/2)*(-1 + sqrt(3)) ~= 5.49 (which is

= (-30 +/- 30*sqrt(3))/4 r=

r = (15/2)*(-1 +/- sqrt(3)), valid only for 0 <r<15

r = (15/2)*(-1 + sqrt(3)) ~= 5.49 (which is in the valid region)

r = 5.49

Explanation: