Question

A positive and a negative charge, each with a magnitude of 3.65 x 10-4 C, are separated by a distance of 0.67 m. Find the force on each of the particles. You must show your work for credit. (Written in GUESS format)

Answer 1

Answer:

The force on each charge is 2668 N and the direction of force on each charge is towards the other charge. Which means force of attraction is there between the charges.

Step-by-Step Explanation:

Let the two charges be q1 and q2

q1 = 3.65*10^(-4) C

q2 = -3.65*10^(-4) C

Distance = d = 0.67 m

We know that the magnitude of the electric force that a particle exerts on another particle is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Also, opposite charges attract each other.

Let F12 is the force on charge 1 by charge 2.

Let F21 is the force on charge 2 by charge 1.

F12 = F21 = F = kq1*q2/d^2     where k = 8.99*10^(9) N m^2 C^(-2)

$= 8.99*10^{9}*(3.65*10^{-4}) *(3.65*10^{-4})/0.67^{2})\\= 8.99*10^{9}*(1.33*10^{-7})/0.67^{2})\\= 1197.7/0.4489\\= 2668 N\\The magnitude of F12 and F21 is 2668 Newton.\\The direction of F12 is towards the charge q2. The direction of F21 is towards the charge q1.$