Question

Find the charge Q that should be placed at the centre of a square of side 6.20 cm with four identical positive charges of 1.40 µC at each corner, so that the whole system is in equilibrium.

Answer 1

take the top left particle then the force exerted on it due to the top right charge is:

F(tr) = k*q^2 / r^2 - to the left ;

where k is 1/(4*Pi*s) I've used s to denote the permittivity of free space, and k is approx = 8.988*10^9Nm^2/c^2. And r = the distance between the charges.

Now due to the bottom left charge it is F(bl) = k*q^2 / r^2 - upwards.

The resultant of these two forces is F(1) = sqrt( F(tr)^2+F(bl)^2 )

F(1) = sqrt( 2(k*q^2 / r^2 )^2) = sqrt(2)*k*q^2 / r^2 - up/left

Now we also have the bottom right charge. If the distance to this charge is R then R = sqrt(r^2+r^2) = sqrt(2)*r

so F(br) = k*q^2 / R^2 = k*q^2 / 2r^2 - up/left

which is the same direction as F(1)
Therefore the overall force on the top teft charge is
F = F(1) + F(br)

F = [sqrt(2)*k*q^2 / r^2] + [k*q^2 / 2r^2]

F = [(2sqrt(2)+1)*k*q^2] / 2r^2

This is the same force experienced by all the q-charges due to the other q charges

Now for Q the force it excerts of the top left q must be of equal magnitude but in the opposite direction. Since it is

at the centre of the square and since the force vectors due to the other q's are pointing away from the centres we

know that Q must be negaitive. Since we know this we can simply find the magnitude of Q by:

Now F(Q) = F

However the distance to Q = 1/2 the distance to the bottom right q; So r(Q) = sqrt(2)*r / 2

k*q*Q / (sqrt(2)*r/2)^2 = [(2sqrt(2)+1)*k*q^2] / 2r^2

Simplifying: 4Q / 2r^2 = (2sqrt(2)+1)*q / 2r^2

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

Q = [2sqrt(2)+1]*(1.40*10^-6)/4 = 1.33*10-6

However we said Q must be negative, so
Q = 1.33uC