Question

A certain sphere is nonconducting. This sphere has six point charges equally spaced from each other fixed to a surface on the sphere.

A) If each point charge has a charge Q, and the radius of the sphere is R, what is the magnitude of the net electric field, E, at one point charge due to the other 5? Please only answer with Q, R and k.

B) Now consider that an actual balloon, when inflated, will exert a force, radially inward on each point charge. Call this force T. At some radius (R0 ) the inward force (T0 ) exactly cancels the electrostatic repulsion of the six point charges, each with a charge of Q0 . This would be a state of static equilibrium between the force of the balloon’s desire to deflate, and the repulsive forces of the point charges on the surface pushing outward. We now decrease the value of these charges to Q2 = Q0/3 which is one third of the original value. The balloon will settle down to a new radius R2 . If the inward force is directly proportional to the radius (T α R3 ), what is the new radius of the sphere (R2 ) where balloon settles?

Answer 1

Solution

A)

The electric field is defined as follows:

Now, if we consider charge three as an interaction point with respect to the other five, we have:

Now, interaction (1) and (3)

The distance between one and other charge is consider as follows:

The distance between the other five charges and one charge is the same because they are equidistant, therefore the repulsive fields are equal, so.

Then

B)

Now, for this case, by inflating the balloon, the net force between each load is obtained as follows:

Therefore

But for ,  

In this condition the radial force in each load balances the electrostatic repulsion force due to each charge, therefore:

Now when the charges decrease to a third of the value q0, the radius varies to R2

Clearing Qo

Substituying

But

Also, To is directly proporcional a R3

Where C is the proportional constant

Clearing R2