Electric Field and Potential: Problem 1.06 & 1.07 with detailed Solutions

1.06) Two equal negative charges -q are kept fixed at points (0, -a) and (0, a) on y-axis. A positive charge +Q is then released from rest at the point (2a, 0) on the x-axis. The point charge +Q will:

A) execute simple harmonic motion about origin

B) move to the origin and remain at rest

C) move to infinity

D) execute oscillatory but not simple harmonic motion

At some intermediate situation, when the charge +Q is at some distance, say x, from the origin:

The electrostatic force between the point charges +Q and -q is attractive in nature with a magnitude F = kQq/r². Also, from the figure shown above we observe that net force acting on the charge +Q is 2Fcosθ directed along the negative x-direction (or directed towards the origin).

Negative sign (-) in the expression of force shows that the resultant force (on +Q) is parallel to the negative x-direction.

From this expression of force, it follows that for any positive displacement (x, measured from the origin) of the particle, net force is directed along negative x-direction and for any negative displacement, the force points along the positive x-direction. In other words, resultant force is always directed towards the origin irrespective of the instantaneous position of the particle. This implies that motion of the particle is oscillatory about origin. However, magnitude of force is not proportional to the displacement (from the mean position or origin) which means it does not represent a simple harmonic motion.

Hence, the particle executes an oscillatory motion but not simple harmonic motion.

1.07) Two infinite linear charges are kept parallel to each other at a separation 0.1 m from each other. If the linear charge density on each is equal to 5 μC/m, then force acting on a unit length of each linear charge is

A) 2.5 N/m

B) 3.25 N/m

C) 4.5 N/m

D) 7.5 N/m

We can assume that one of the two linear charges is placed in the electric field of the other. The field generated by a long linear charge at a distance d from it has a magnitude 2kλ/d where λ is the charge density. On substituting x = 0.1 m and λ = 5 μC we get

This much strength of electric field is produced by one of the linear charges at a distance of 0.1 m where the other linear charge is present. So force acting on the unit length of that other linear charge = λE (as λ represents charge per unit length). 

∴ Force per unit length = λE = (5×10⁻⁶)(9×10⁵) N/m = 4.5 N/m. This force is repulsive in nature as both the linear charges are positive (likely charged).