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.
