Question

A point charge is projected along the axis of circular ring of charge Q and radius 102–√ cm. The distance of the point charge from centre of ring, where acceleration of charged particle is maximum, will be

Answer 1

Acceleration of charged particle depends upon electric field intensity of region.
because acceleration = qE/m , here q is charge , E is electric field intensity and m is the mass of particle. hence, higher the electric field higher will be acceleration of particle. So, we have to find out the point where electric field will be maximum.

we know, elecric field on the axis of ring is given by
E = kqx/(r² + x²)^3/2 , here x is the distance between centre of ring to observation point on the axis of ring, r is the radius of ring and q is the charged particle.
For finding maximum electric field differentiate E with respect to x
dE/dx = kq[(r² + x²)^3/2 - 3/2(r² + x²)½ 2x² ]/(r² + x²)³
0 = (r² + x²)½ [ r² + x² - 3x² ]/(r² + x²)³
r² = 2x² ⇒r = ±√2x
Hence, x = ±r/√2
Check d²E/dx² at x = r/√2 , you will get d²E/dx² < 0
Hence, at x = r/√2 , E will be maximum .
So, E will be maximum at x = 10√2/√2 = 10cm
Hence, acceleration will be maximum at x = 10cm

Answer is 10cm