Four point charges, q , are fixed to the four corners of a square that is 11.2 c

Question

Four point charges, q, are fixed to the four corners of a square that is 11.2 cm on a side. An electron is suspended above a point at which its weight is balanced by the electrostatic force due to the four point charges, at a distance of 11 nm above the center of the square. (The square is horizontally flat, and the electron is suspended 11 nm vertically above the center of the square.) What is the magnitude of each fixed charge in coulombs?
                      C

What is the magnitude of each fixed charge as a multiple of the electron's charge?
                      e

Explanation / Answer

The length of the square is L = 11.2 cm

The distance of the suspension of electron is d = 12 nm

The mass of the electron is me = 9.11*10-31 kg

The charge of the electron e = 1.6*10-19 C

The maginitude of the each charge at the cormer be q

The distance of the half diagonal of the square is l = 0.5v[(11.2*10-2 m)2+(11.2*10-2 m)2]
l = 0.1668 m

The distance of the electron from each corner of the square is
L = v[(0.1668 m)2+(12*10-9 m)2] = 0.1668 m

The angle made by the line joining the electro and the charge is ? = tan-1(12*10-9 m/0.1668 m)
? = 4.122*10-6

The electric force between two charges of magnitudes q1 and q2 seperated by distance d is given by

F = kq1q2/d2
k = 9*109 Nm2/C2 is the coulomb's constant

All the charges are similar, therefore, the horizontal components are get cancelled and the vertical components are added.

F = 4*(9*109 Nm2/C2 )(q)(1.6*10-19 C)(sin4.122*10-6)/(0.1668 m)2 (four similar charges)

F = q*1.486*10-14 N

As this force can balance the electron,

F = meg

g is the acceleration due to geavity

F = (9.11*10-31 kg)(9.8 m/s2)

F = 8.9278*10-30 N

Therefore

q*1.489*10-14 N = 8.9278*10-30 N

q = 5.99*10-16 C

The number of electrons equivalant to this charge is

N = 5.99*10-16 C / 1.6*10-19 C

N = 3747.4 eletrons

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