Electrostatics Doubt (electric field)

[sudden hazel]

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[toxic lark]

I tried solving and brainstomring on this but nothing worked , give me some hints or steps of approach i'll try on my own

[autumn cradle]

You can equate magnitude of electric field by a point charge and magnitude of electric field given for a point like (R,0)

Solution 👇

If you have an answer then tell me if im correct or not

[pine hollow]

but this EF isn't due to the sphere.

[autumn cradle]

Its due to charge

[toxic lark]

yea this is right

[autumn cradle]

Do You understood the solution?

[waxen lake]

The motive behind this problem is to think about electric field due to infinite line charge.
Electric field can be written as alpha/r (r^) ( polar coordinate)
on comparing it with electric field due to infinite line charge you will get lambda=2(pi)(epsilon)(alpha)
Now consider an infinite line charge lying along z axis and it cuts the sphere than the length of that line inside sphere will be 2R so Q=lambda(2R) =4(pi)(epsilon)R(alpha)

[autumn cradle]

I did the exact same thing for a plane and you did by assuming 3D

[waxen lake]

you cannot consider the electric field due to a point charge as for that the field should be given by E= alpha (x i + y j) / (x^2+y^2)^3/2
because it reduces to alpha (x i + y j)/root(x^2+y^2) * (1/x^2+y^2)
that is alpha/r^2 (r^) similar to electric field due to point charge

[autumn cradle]

Can you write it on paper i can't understand the text form 😅

[autumn cradle]

Ok ok then what is the flaw in my solution

[waxen lake]

You said that you assumed a charged particle placed at (0,0,0) and you equated the electric field due to that charged particle and the given electric field. From there you get the charge of the particle that you assumed to be placed at (0,0,0).Now as you are saying that the charge you calculated is equal to the charge of sphere basically you said that the charge placed at (0,0,0) is the charge that the sphere possess.So ultimately you are saying that the given electric field in question was of sphere and we need to calculate the charge possessed by it. And from there using shell theorem you assumed all charge at center and equated the electric field due to it and the given electric field .Now my question is, is it necessary that the given electric field is caused by the sphere only ? can't there be any external electric field existing in space in the given situation.
Like in my solution you can see i assumed the field produced due to infinite line charge kept along some axis (say z) and the portion of the line inside the sphere was the charge possessed.

[autumn cradle]

Ohk I got it. I misunderstood the language and thought that both charge and circle have centre at origin. Thanks for correctting me.
I came up with another solution by using gauss's law and Concept of flux un non uniform electric field...

@toxic lark @waxen lake

[toxic lark]

W, thank you both @waxen lake @autumn cradle

[waxen lake]

It would have been better if you assumed an arbitrary point in space and for that area vector direction would be along r^ and field direction would also have been along r^ and then writing gauss law.
Basically you fixed the y coordinate as per your first line so idk, like if there is an integration involving two variable and you assumed y=0 and then integrating it ,i don't find it correct as while integrating you should chose an arbitrary point rather than some point (say R,0).

[autumn cradle]

Bro i couldn't take any arbitrary point for area vector in SPACE i could only take the points on the SPHERE. Thats how gauss's law works and putting any point on sphere then always x²+y² = R² and dot product of the area vector an Electric field will always be R as [(xî + yĵ).(xî + yĵ)]÷[(x²+y²)(x²+y²)^½] = R²^½ = R

I think i am absolutely correct in my position

[waxen lake]

From arbitrary point in space i meant an arbitrary point on the sphere (not (R,0)).
but no point of discussion "if you are absolutely correct" .

[autumn cradle]

Bro i also mentioned the arbitrary condition which you are saying is also true. I only said that taking (R,0) is not wrong as it satisfies the eqution of sphere and pls be nice not sarcastic and chill bro galti sabse ho sakti hai mere se bhi hui thi na phli baar me

[autumn cradle]

@toxic lark i think your doubt is solved

[toxic lark]

yep thanks guys

+solved @autumn cradle