IIT JEE , AIEEE Forum - Physics , Electricity

pinnaka

After A request made by Indian army to solve this Question. But I found that the Question was not stated properly
But Now I have a got a copy of iit paper the Question is

Q) A sphericsl portion has been removed from a solid sphere having charge uniformly disturburted through out the volume. The E inside the emptied space is -----

The ans is it is uniform and non zero

NO it is given in the Q that the charge is distributed uniformly throughout the entire volume therfore the sphere is not a conducting sphere. If it were to be the enitre charge ould reside on the surface

Here is the mathematical proof
The E of a solid sphere of uniform density is Qr/4*pie*R^3
E = rho*r/3eplison_0
E(vector) = rho*r(vector)/ 3*eplison_0-----
Now assume that the removed portion is a disatnce b (center of cavity) from the center of the sphere
A point inside the sphere is located by r-b (vector from)

---Hope taht helps
E = E(uniform) +E(hole)
E unifrom represents the sphere of uniform charge rho and E(hole) of charge density -rho
inside the cavity net charge is zero
E(uniform) = rho*r/3*eplison_0
E_hole = - rho*(r-b) {vector form)/3*eplison_0
then E = pb(vect0r)/3*eplison_0
Therfore this magnitude is consant and non zero

therefore the ans non zero and constant electric filed

anchal_1991

and gauss is valid for large surfaces

anchal_1991

electric field in assymetric caity is (kq)/(R²+r²+Rr)

if r is radius of cavity
nd R is radius of sphere

electric field is 0 for symmetric cavity

sohil1989

Everybody,

The flux inside is zero as many of you suggested but it deos not necessarily mean Electric Field is zero.

Confused, here is a simple example.

Draw a sphere a bit away from the point charge so that it does not enclose it.

On applying guass theorem we can say that flux through the sphere is zero but we all know that Electric Field is not.

The mistake all are commiting is

E.dS=

=0 does not mean E

dS = 0.

We can do this only incase of symetrical bodies where E is constant but it is not the case here.Since E is variable it cannot be taken outand is not Zero.

Even fitjee and solution from other sources is wrong becoz if we assume E to be constantit has to be Zero so answer is c.

I am right.Just think about it.

aditya_arora04

Hi,

Maybe i can add a little to sidlol.

See, from all sides of the cavity, At one side charge is more, and at other side charge is less, due to which, the net electric flux is not 0 and not uniform.

akhilprem

Okay, so u guys know that there is field present. I'm pretty sure that it is non uniform. Here's my logic:

The net flux inside the cavity is zero quoting resnick and halliday. I'm familiar with the text (Oh I just love that book!). But net flux can only be zero in 2 cases:

1) No lines of force enter the cavity

2) The same number of lines that enter the cavity also leave it

Now (1) is not possible here because the sphere is non conducting and the charges stay fixed in their positions os thet there is an excess of charges on one side due to lack of symmetry.

However (2) IS possible. The lines of force would gradually bent around and leave the cavity. Due to lack to spherical symmetry the lines of force could be curved and spaced to different extent (remember the images from resnick and halliday) which means field would be non uniform.

akhilprem

pinakka's reasoning seems to be wrong since the formula he used is applicable only to points inside a uniformly charged non conducting sphere. It cannot be applied to a point within a cavity inside the sphere.

robinphilip1989

this is not a conducting sphere why should the charge in the cavity be zero
besides applying gauss law is not advised as the system is not symmetrical
u can solve the problem by superposition assume an oppositely charged sphere in the cavity and take any point inside it. assume then the whole sphere to be charged as we have placed an opp charged body

vectorially write the forces

there was a similar question asked in the mains before on gravitation u can go thru that for calculations

akhilprem

Gauss law holds for any shape of the charge distribution. Could u explain ur answer in more detail.

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