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A positively charged rod lies along X-ax...

A positively charged rod lies along X-axis in such a manner that one end of the rod is at the origin and the other end at `x =-oo`. The linear charge density for the rod is `lambda`. Electric field intensity at x =a is

A

`vecE=lamda/(4piepsilon_0a)(-hati)`

B

`vecE=lamda/(4piepsilon_0a)(-hati)`

C

`vecE=lamda/(2piepsilon_0a)(-hati)`

D

`vecE=lamda/(2epiepsilon_0a)(-hati)`

Text Solution

Verified by Experts

The correct Answer is:
A

Charge on dx is `dq = lamdadx`
Electric field due to element dx
`dE= (dq)/(4piepsilon_0x^2)implies E=int_a^(oo)(lamdadx)/(4piepsilon_0x^2)`
`implies E= (lamda)/(4piepsilon_0)[(x^(-2+1))/(-2+1)]_a^(oo) =(lamda)/(4piepsilon_0a)implies vecE =(lamda)/(4piepsilon_0a)(-hati)`
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