Question

six equal positive charges,q , are placed at the vertices of regular hexagon of side a. the charge kept at the centre of the hexagon for the equilibrium of system is
a -q b -q/6 c 1.83q d -1.83q

Answer 1

I'd go with answer a. If the center charge is -q then the net force on that charge will be 0 so the system will be at equilibrium.

Answer 2

It should be true for all a,b,c,d .. whatever be the charge placed at the centre it will experience no force since net electric field at centre will be zero.. The centre is a null point for that system..

Answer 3

Nice problem. The negative charge is stable no matter what, since it is equally attracted to all six positive charges. The challenge is to figure out how large the negative charge needs to be to hold all six positive charges against their mutual repulsion.

Because of symmetry, you can just calculate the net force along the axis joining one of the positive charges with the negative charge. As long as that is zero, the system is stable.

The magnitudes of the forces are easy to calculate through Coulomb's Law and geometry. Then you need to resolve each force into components parallel and perpendicular to the chosen axis, which because of the geometry involved, is pretty easy.

When I do this, setting the charge on the negative charge to -fq, I get the following equation:

- fq^2/a^2 + 2 (q^2/a^2) cos(60) + 2 (q^2/(3a)^2) sin(60) + q^2/(2a)^2 = 0

The terms here are, in order, the attraction from the central negative charge, the repulsion from the two closest positive neighbors, the repulsion from the two positive neighbors across the hexagon at an angle, and the repulsion from the positive charge directly across the hexagon. Notice the sines and cosines to find the component of each force along the axis between my positive charge and the central negative charge.

q^2 and a^2 can be divided out, and solving for f gives you:

f = 1 + 1/4 + 1/sqrt(3) = 1.82735...

So the answer is (d).