Question

Two small nonconducting spheres have a total charge of Q=Q1+Q2= 90.0 μC , Q1<Q2. When placed 31.0 cm apart, the force each exerts on the other is 13.0 N and is repulsive.

Part A: What is the charge Q1?

Part B: What is the charge Q2?

Part C: What would Q1 be if the force were attractive?

Part D: What would Q2 be if the force were attractive?

Answer 1

●Given:

Q=Q₁+Q^{​​​​​​}₂=90.0 × 10^{-6 C}

r = 31 cm = 31 × 10^-2 m

F = 13.0 N

Let's say charge on one sphere is A, then the charge on the other sphere mathematically should be 90.0-A.

Now, we use Coulomb's law and derive an expression for the force between the two spheres.

  

  

[You may wonder from where did I get 10^-12 from. Well, if you think about it, charge A will be in microcoulombs which is basically 10^-6 coulombs and 90.0-A will also be in 10^-6 coulombs. So when you multiply them you get 10^-12 C and just shift it to the other side so that you can have values on one side and expression on the other.]

Now solve the quadratic equation,

A² - 90A + 139 = 0

We get,

A = 88.43 and A = 1.57

Since, Q₁<Q₂

A. Q₁ = 1.57  

B. Q₂ = 88.43

Now, if the force is attractive, follow the same steps as above with a change in the sign of F.

ie, use -13.0 N instead of 13.0 N

Thus, we get another quadratic equation of the form,

A² - 90A - 139 = 0

Solving this, we get,

A = 91.5 and A = - 1.5

So,

C.  Q₁ = - 1.5

D.  Q₂ = 91.5