Question

a charge Q is distributed uniformly on a fixed ring of radius a .A point charge q is placed on axis at a distance 2a from the centre of ring. The force experienced by point charge at given position is

Answer 1

(a) if q > 0 and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre.

(b) if q < 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring.

(c) if q < 0, it will perform SHM for small displacement along the axis.

(d) q at the centre of the ring is in an unstable equilibrium within the plane of the ring for q > 0.

F =kOq/R^2

Answer 2

Answer:

The force experienced by the charge q is = $\frac{KqQ 2a}{(5^\frac{3}{2}a^3)}$

Explanation:

• Electric field ‘E’ at ‘P’ due to the charged ring
  $E = \frac{KQx}{(R^2+x^2)^\frac{3}{2}}$
• Now here we know that point p is at a distance of 2a from the center of the ring of radius a so the value of the Electric feild E is
  $E = \frac{KQ2a}{((2a)^2+a^2)^\frac{3}{2}}$
• Now as we know that force on the charge Q placed at a distance from the center is $F = Q*E$
• Now here value of Q is q and value of electric feild E is $E = \frac{KQ2a}{((2a)^2+a^2)^\frac{3}{2}}$
• Hence the value of Force F is
  $F=\frac{KqQ 2a}{(5^\frac{3}{2}a^3)}$
  #SPJ2