Question

Four charges are kept at the corners of a square of side a.A charge q is placed at the centre. Get the relation between Q and q if the whole syatem is in equillibrium?

Answer 1

Let us call the square as ABCD and its centre as O. Further, AB=BC=CD=DA= say a metres.

For any corner of the square, say A:-

The resultant force on the charge +q placed @ A is the vector sum of the repulsion to q at A from the charges at b, c and D.

This resultant is directed radially outward along the extension of the diagonal CA and has the magnitude (1/4πε0){(q^2)/(a^2)} {[(2)^(1/2)] + [1/2]} ------------------------------> (1)

The above also holds for the charges placed at B or C or D because of the symmetry.

Let us now place a charge x at the centre O to maintain equilibrium among the charges at A, B, C, D. 

This should exert an equal and opposite force to the outward force shown in (1) above.

So it needs to be a negative charge. The distance OA, OB, OC, OD from the respective corners A, B, C or D is half of the diagonals (AC or BD) i.e. (1/2){(a)[2^(1/2)]} i.e. {(a)/[(2^(1/2)]}.

Attractive force on the charge at A (or B or C or D) due to the charge x at O = (1/4πε0)(qx)/{{(a)/[(2^(1/2)]}^(2)} = 2(qx)/{(a)^(2)} ------------------------------> (2)

Equating (1) and (2), we get x = (q/2){[(2)^(1/2)] + [1/2]} and a negative charge.