Question

Four identical charges, Q, occupy the corners of a square with sides of length a. A fifth charge, q, can be placed at any desired location. Part A Find the location of the fifth charge.

Answer 1

Answer:

  q = - ( 2*sqrt(2) + 1 )*Q / 4

Explanation:

Given:

- Side of each square L = a

- Charge q is placed among 4 other identical charges Q.

Find:

- Find the location and magnitude of the fifth charge q such that net Electric Force at its position is zero.

Solution:

- Compute distance r  from charge Q to q i.e center of all four charges Q:

                          r = 0.5*sqrt ( a^2 + a^2 )

                          r = a / 2sqrt(2)

- Compute the individual Electrostatic forces @ point A:

                          F_b = F_d = k*Q^2 / a^2

                          F_c = k*Q^2 / (a*sqrt(2))^2 = k*Q^2 / 2*a^2

                          F = k*Q*q / (a / 2sqrt(2))^2 = 2*k*Q^2 / a^2

- Use Electrostatic Equilibrium conditions:

                          F_b + F_c*cos(45) = F*cos(45)

                          F_d + F_c*sin(45) = F*sin(45)

- Plug in the values and equate:

                          { (Q/a)^2 + (Q^2 / 2*a^2*sqrt(2)) = sqrt(2)*Q*q / a^2

- Canceling all k's and a^2:

                          Q * ( 1 + (1 /2*sqrt(2)) ) =  sqrt(2)*q

                          q = - ( 2*sqrt(2) + 1 )*Q / 4

- Note: In attachment Q's and q's are interchange but the solution here provided is according to the question at hand.