# Question #ffba3

Dec 16, 2016

The answer as posted is 'the result of the calculation is zero'.

This is due to symmetry of the charge distribution around the charge $q$ placed at the center. Let's see the figure below

Line charge density of charge $Q$ spread along the circumference of the circle is

${\rho}_{L} = \frac{Q}{2 \pi R}$ ....(1)

Consider any length element $\mathrm{dl}$ on the circumference. Charge on this element

$\mathrm{dQ} = \mathrm{dl} \times \frac{Q}{2 \pi R}$ ....(2)

Force between this charge element and charge $q$ placed at the center is given by Coulomb's law

$| \vec{F} | = k \frac{q \times \mathrm{dQ}}{R} ^ 2$ .....(3)

The direction of force will be along the radius and will be towards or away from the center depending upon the sign of charges.

Now locate another length element $\mathrm{dl}$ on the circumference located diametrically opposite.
Charge on this element is also $= \mathrm{dQ}$

Force between this charge element and charge $q$ placed at the center is same as given by equation (3)

However, the direction of force will be opposite to the direction of the force exerted by first charge element.

Hence, net force on the charge $q$ due to the pair of charge elements $= 0$.
It is seen that the charge distribution can be considered to be made up of infinite numbers of paired charge elements, net force due to each pair being $= 0$.

Final result follows.