Question

Particle A has charge qA and particle B has charge qB. When they are separated by a distance ri, they experience an attractive force Fi. The particles are moved without altering their charges. Now they experience an attractive force with a magnitude of 36Fi. Find an expression for their new separation.

Answer 1

`r_f=(1)/(6)r_i`

Step-by-step explanation:

To find the new separation of the charges, you first take into account the formula for the electric force, when the force are separated a distance of ri.

You use the following expression:

`F_i=k(q_Aq_B)/(r_i^2)` (1)

k: Coulomb's constant

qA: charge of A particle

qB: charge of B particle

When the charges are separated to a new distance rf, the new force is 36Fi, if the charges have not changed, you have:

`F_f=36F_i=k(q_Aq_B)/(r_f^2)` (2)

To find the new separation you replace the expression for Fi of the equation (1) into the equation (2) and solve for rf in terms of ri:

`36F_i=36k(q_Aq_B)/(r_i^2)=k(q_Aq_B)/(r_f^2)\\\\(36)/(r_i^2)=(1)/(r_f^2)\\\\r_f=(1)/(6)r_i`

The new separation of the charges is 1/6 times of the initial separation