Question

Two small nonconducting spheres have a total charge of 86.9μC. When placed 1.11 m apart, the force each exerts on the other is 11.7 N and is repulsive. What is the (larger) charge on the spheres? Incorrect. Tries 1/8 Previous Tries What is the (smaller) charge on the other sphere? Tries 0/8 What if the force were attractive? What is the larger charge on one of the spheres? Tries 0/8 What is the smaller charge on the sphere?

Answer 1

let’s say charge on one sphere is A

then the charge on the other sphere mathematically should be 86.9-A

now we use coulomb’s law and derive an expression for the force between the two spheres

F=KA(86.9-A) / r²

11.7=(9x10^9)(A)(86.9-A) / 1.11²

(11.7)(1.11)² / (9x10^9) (10^-12) = A(86.9-A)

you may wonder from where did we get 10^-12 from. Well if you think about it Charge A will be in microcoulombs which is basically 10^-6 coloumbs and 86.9-A will also be in 10^-6 coloumbs soo when you multiply them you get 10^-12 C and just shift it to the other side

so that you have values on one side and expression on the other

Now lets simplify it,

1602 = 86.9A-A²

now solve it as a quadratic equation

A²-86.9A + 1602 =0

A=60.36μC and A=26.54μC

if you add them you get 86.9μC

(a) so the larger is 60.4μC

(b) smaller is 26.5μC

you can do the C and D part yourself now

just change 11.7 into -11.7

Our final equation will become

A²-86.9A-1602=0, we get:

A=102.5μC and A=-15.6μC

(c) so the larger is 102.5μC

(d) smaller is -15.6μC