IIT JEE , AIEEE Forum - Physics , Electricity

shinee

A rod AB of length L has a charge per unit length

=Cx where x is the distance of the point on the rod from the end A. Find the electric field on the axis of the rod at a diatance r from
a) A
b) B
i am unable to workout with the integration . can u plz help
Salutes assured

karthik2007

I'll help you out with the first one. The second part follows on the same lines:

Consider an element of the rod enclosed between the lengths x and x+dx from end A.

Charge on this element (We consider an element as Coloumb's law is valid only for point charges) = C(x+dx) - Cx = Cdx

Now, the electric field due to this element at Point P is:

dE = kdq/(r+x)²

or dE = kCdx/(r+x)²

Hence E = _{[0 ]}

^{[L ]} kCdx/(r+x)²

Now you may carry on :)

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