F5 PHY ELECTRICITY

1.Let d be the distance from the 9 nC charge that a neutral point is formed.
At the neutral point, the electric field is zero
Hence, k(9)/d^2 + k(-1)/(d+0.1)^2 = 0
i.e. k(9)/d^2 = k/(d+0.1)^2
simplifying, 8d^2 + 1.8d + 0.09 = 0
solve for d gives d = -0.075 m or - 0.15 m

Since the neutral point would not exist in between the two charges, hence the only neutral point is at a distance of 0.15 m (or 15 cm) to the rigth hand side of the 9 nC charge.

2. (c). The net force on a free electron at M is zero. If not, it will move until equilibrium is reached.

Hence, the force given by the charge q on the free electron must be balanced by the +ve and -ve induced charges at the two ends of the rod.

Force on electron by charge q = k(q)e/(d+l/2)^2
where k is a constant (= 9x10^9 F/m)
e is the electronic charge (=1.6 x 10^-19 C)
Hence, force on electron given by the induced charges = (1.44x10^-9)q/(d+l/2)^2, and acts in the direction towards the end B of the rod.