Saved Bookmarks
| 1. |
A charge is distributed uniformaly over is 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. |
|
Answer» Solution :Consider a UNIFORM circular ring of radius .a. carrying a charge Q distributed uniformaly over its surface Let P be a point situated at a distance r from the centre of ring along its axis. Consider an element of length Dl around point A of the ring carrying a charge Dq `=(Q)/(2 pi A) Delta l ` The electrical field at point P due to this element is given as ` ""| oversetto (Delta E) |=(1)/( 4 pi in _0).(Delta q)/((AP)^(2))=(1)/( 4 pi in _0).(Q)/(2 pi a ( a^(2) +r^(2)) ` The ELECTRIC field ` oversetto (Delta E )`, directed along the direction AP, subtends an angle ` theta ` with the axis of ring and can be resolved into TWO components namely (i)`Delta Ecos theta ` along the axis of ring , and (ii)` Delta Esin theta ` normal to the axis of ring. It is clear fromsymmetry that the normal components `Delta E sin theta ` due to mutuallyopposite charges elements at A and B nullify each other. Hence , net electricfield due to whole ring will be ` E= sum Delta E cos theta =sum ((1)/(4 pi in _0) .(QDelta l )/(2 pi a (a^(2) +r^(2)) ). (r)/( (a^(2) +r^(2))^(1//2)) ) ` ` =(1)/(4 pi in_0).(Qr)/( 2pi a (r^(2) +a^(2))^(3//2)) .sum Delta l =(1)/( 4 pi in _0) . (Qr)/( 4 pi a(r^(2)+a^(2) )^(3//2)). 2pi a =(1)/( 4 pi in _0) .(Qr)/( (r^(2) +a^(2)) ^(3//2))` The field `oversetto E ` is directed along the axisOP of the charged ring. If ` gtgt a,`then the above relation may be expressed as ` oversetto E= (1)/( 4 pi in _0) .(Q)/(r^(2)) hat r ` This shows that for far POINTS at long distances from the ring. it behaves lika a point charges. ` (##U_LIK_SP_PHY_XII_C01_E10_001_S01.png" width="80%"> |
|