Question

a spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume. the electric field inside the emptied space is
0 everywhere
non uniform
non 0 and uniform
zero only at its centre

Answer 1

0 everywhere (gauss law)

Answer 2

Using Gauß's law twice, you end up with a non 0 field.

Answer 3

how??

Answer 4

Apply Gauß's law to a the full bigger sphere (centre O) uniformly charged \(\rho \).

Apply again to smaller sphere (centre A) uniformly charged \(-\rho \).

Add up fields.

Answer 5

i dunno gaub's law only gauss'...
But at first intuitively i also thought that it should be non zero as it doesnt say if it a conductor or not...but if E is non zero then graphically there will be lines of force inside which originate from positive but ends at negative..since the line have nowhere to end inside(no negative charge inside),therefore there must not be any lines inside at all and hence E must be zero everywhere,wouldn't it??

Answer 6

sry, i've learnt only gauss' law so dont know anything about your argument :(
but please point where i m wrong..

Answer 7

sry gotta go..willl ask later. and thanks for telling me about gaub's law.i thought EM was over after maxwell's eqns..glad to know there is some more nice things to study :)

Answer 8

ß is German's double s, so we are both talking about the same law :-))

Sphere cannot be a conductor because a conductor does not admit volume charges.

Using superposition theorem:

For point M inside the whole sphere (alone):

\( \vec E_1(M) = \Large \frac{\rho}{3\epsilon_o}\normalsize \vec {OM}\)

For point M inside the smaller sphere (alone):

\( \vec E_2(M) =- \Large \frac{\rho}{3\epsilon_o}\normalsize \vec {AM}\)

Total field:

\( \vec E(M) = \Large \frac{\rho}{3\epsilon_o}\normalsize (\vec {OM}-\vec {AM})=\Large \frac{\rho}{3\epsilon_o}\normalsize \vec {OA}\)

So field is uniform and non-zero.

Answer 9

Oh..honest mistake :) never knew German....but for that argument,what about the electric lines of force??how will they look like inside the sphere?

Answer 10

Parallel to line OA

Answer 11

but are we not just making up the negatively charged sphere for problem solving using superposition?after all it doesn't have any physical existence.if we look at it from the properties of ELOF's then by:
1.cant be discontinuous
2.always ends at negative charge and start on positive charge
if these two are to be satisfied then where will the lines end(escape) if they are inside the spherical cavity of the sphere..

Answer 12

They will leave the cavity , cross the remaining part of the solid sphere and go to infinity.

I do not get your point about superposition.

Answer 13

unless i misunderstood your explanation,i think you said that to find the elecric field at a point we consider the whole sphere of + charge and superpose a - sphere covering the cavity.and then find the E by superposition of E field by both spheres. right?

Answer 14

Right!

Answer 15

where will these lines inside go??...they cant go through the sphere as elofs cant pass through charged sphere(unless the field is very strong) and they can only start from the inner surface of spheres...