Q11 of 26 Page 1

A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.

Electrical Intensity ( for a point charge at a distance from the ring


A ring is placed in this experiment for the finding of electrical intensity is of radius r, and a uniformly positive charge is distributed on the ring.



The charge density () on the above ring is



Due to perpendicular charge density on them the electrical components cancels each other out.


Therefore, the magnitude of electrical field is





where


is the Electrical Intensity, is the charge, is the distance between ring centre and the charge, a is the radius of the ring.


Now


a). If the distance x = 0, then the intensity is


Electrical Intensity = 0


b). If the distance x>>a, then the intensity is


Electrical Intensity =


More from this chapter

All 26 →
9

Two cells of emfs 1.5 V and 2.0 V having internal resistances 0.2 Ω and 0.3 Ω respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.

 10

State Brewster’s law. The value of Brewster angle for a transparent medium is different for light of different colours. Give reason.

 12

Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation.

 13

(a) Write the expression for the magnetic force acting on a charged particle moving with velocity n in the presence of magnetic field B.

(b) A neutron, an electron and an alpha particle moving with equal velocities, enter a uniform magnetic field going into the plane of the paper as shown. Trace their paths in the field and justify your answer.