1.

A very long uniformly charged thin wirewhose one end is at point (0,a) extends along positive y-axis . The charge per unit length of wire is lamda . The electric field at origin is

Answer»

`(lamda)/(4piepsi_0a)(-hatj)`
`(lamda)/(2piepsi_0a)(hatj)`
`(lamda)/(4piepsi_0a)(-hati-hatj)`
`(lamda)/(2piepsi_0a)(hati)`

Solution :Use `dE = (Kdq)/(r^2), " where " K=(1)/(4piepsi_0)`

Consider on INFINITESIMALLY small element of length dy at a distance y from origin O. The CHARGE on this section is `dq=lamdady`
Now, `dvecE_0 = (1)/(4piepsi_0).(dq)/(y^2) (-hatj)`
`vecE_0 = (lamda)/(4piepsi_0) underset(a) OVERSET(00)INT (dy)/(y^2)(-hatj)= (-lamdahatj)/(4piepsi_0)[(1)/(-y)]_a^(oo)`
`=(-lamdahatj)/(4piepsi_0) [ - (1)/(oo)- (-1/a)]`
`VECE=(lamda)/(4piepsi_0a)(-hatj)`


Discussion

No Comment Found

Related InterviewSolutions