A very long straight uniformly charged thread carries a charge ?(LAMBDA) per unit length.Find the magnitude...

jasbir

and direction of the electric field strength at a point which is at a distance 'y' from the thread and lies on the perpendicular passing through one of the thread's end.

I HAVE TRIED THIS QUESTION BY APPLYING GAUSS'S THEOREM BUT I M NOT GETTING THE CORRECT ANSWER AND ALSO I DON'T KNOW HOW TO FIND THE DIRECTION.THE ANSWERS IS
E=[k?(2)^1/2]/y
The vector E is directed at the angle 45 to the thread.
plzz solve my problem
its I.E.IRODOV'S problem number 3.14

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^:raised to the power

sagarvaze

you cannot apply gauss law to this problem as there is no symmetry you have to consider a small piece of length dx at distance x from the end and then find the field due to it at that point then resolve it into components then integrate them from 0 to infinity separately then again find equivalent vector.

karthik2007

karthik2007

Hope you noted that dEx = dE x cos@ and alike for the vertical component... I didnt mention it.

jasbir

yarr
i didn't understood which angle is 'a' and how did u set the limits from 0 to 90

plzz explain...

jasbir

and plzz explain why there is no symmetry...

karthik2007

Dear jasbir:

We have to basically evaluate the integral

kLy _{[0 ]}

^{[infinity ]} dx/(x²+y²)^3/2

to solve this integral, we have to use the substitution
x = ytana

where a is any arbitrary angle... its just used to simplify the integral.

Also, to change the limits:

When x= infinity, tana = infinity , hence, a = pi/2 (As tan^-1infinity = pi/2)

When x = infinity, tan a = 0, hence a = 0.

(Where tan a = x/y)

karthik2007

any more problems jasbir?

studyid

well ....... @ karthik :

brilliant method ..........

but i wud like to mention ...that ......last year .....when i had been to the

bt programme ..... the proff. there solved this very same sum by the

Gauss Theorem ............. !!!!!

karthik2007

Well... I thought over it, but ended up with this:

E.ds = Q_in/

o

But Q_in = infinity :(

karthik2007

I probably need to open up my mind to think of a better guassian surface... I'll think over this one. You've given me something to ponder about studyid :)

studyid

ya sure .......... i am preparing the method .........

just wait a few minutes .........i am doing the needful .....

studyid

Well here is the method :

Let us suppose the gaussian surface as shown ...

now the direction electric field due to the line charge will be towards the right ....

making some angle theta with the horizontal ....

compare the the first figure with the second figure ..

by comparison .......

we see ....

that

2Ecos (theta) = lamba / 2*Pi*Epsilon₀ *y

and by symmetry .. theta = 45 degrees

hence .......

E = lamba / 2*root(2)*Pi*Epsilon₀ *y .....

karthik2007

Nice... nice method? You thought of this?

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A very long straight uniformly charged thread carries a charge ?(LAMBDA) per unit length.Find the magnitude...

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