Question

Two similarly charged spheres A and B separated by a distance 'r' with force of 2×10^-5N. A third identical uncharged sphere C is touched to A and then placed between a mid point of A and B. Calculate net electrostatic force on C?

Answer 1

`"Please have a look at the fallowing explanation."`

Explanation:

• make a drawing to facilitate the solution

• Coulomb's law can help you calculate the force that two electric charges apply to each other.

• `F=k*(q_1*q_2)/r^2`

• `color(red)(2*10^-5)=k*(q*q)/r^2=color(red)(k*q^2/r^2)`

• The force is directly proportional to the product of the electric charges.
• The force is inversely proportional to the square of the distance between the electric charges.

• Since the spheres are identical, the total load is shared equally

• The new electric charge distribution of the A and C spheres is shown in Figure 3.

• Figure 4 shows the new load sequence.

• C is pushed by both A and B.

• `F_1=k*(q/2*q/2)/(r/2)^2=k*(q^2/4)/(r^2/4)=color(red)(k*q^2/r^2=2.10^-5)`

• `F_2=k*(q*q/2)/(r/2)^2=k*(q^2/2)/(r^2/4)=2color(red)(k*q^2/4)=color(red)(2*2*10^-5=4*10^-5)`

• Now we must find the vector sum of `F_1""` and`" " F_2`.

• `vec R=vec F_1 +vec F_2`

• magnitude of net electrostatic force over C is:

• `-4*10^-5+2*10^-5=-2*10^-5" Newtons"`

• The direction to B is considered positive.
  -The direction of net electrostatic force is toward A.