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Electricity

New kid on the Block

 Joined: 18 Aug 2009 Post: 19
18 Aug 2009 19:20:11 IST
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Electric field inside a cavity of a non conducting uniform sphere is constant.Can anyone pls explain
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Electric field inside a cavity of a non conducting uniform sphere is constant.Can anyone pls explain me why is it uniform and if it is uniform what is its value??

Scorching goIITian

Joined: 4 Apr 2009
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18 Aug 2009 19:25:59 IST
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frm gauss law the flux=integral E.dS = q/epsilon, so E= Q/4PI epsilon r^2

Blazing goIITian

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18 Aug 2009 19:28:11 IST
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New kid on the Block

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18 Aug 2009 19:30:45 IST
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@karma Answer is nt 0 for sure.@sravanthi who said the charge enclosed is Q....???

New kid on the Block

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19 Aug 2009 16:26:25 IST
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sum1 pls help!!!!!

Blazing goIITian

Joined: 8 Oct 2008
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19 Aug 2009 20:08:17 IST
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hello dear

apply the guass theorem to solve this question

according to the guass theorem

E.A = (charge enclosed) / (epsolon)

here charge enclosed is zero hence  E = zero

New kid on the Block

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19 Aug 2009 22:21:59 IST
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even i thought the same thing before but IIT 2007 had asked a question on the same concpt ..

A spherical portion has been removed from solid sphere having charge uniformly distributed in its volume as shown in the figure.The electric field inside the emptied space is

(A) zero everywhere           (B)non zero and uniform      (C) non uniform          (D) zero only at its centre.

Answer is option B....I checked out many solutions but the answer is B..even i am not understanding why....pls help me someone.....

Scorching goIITian

Joined: 4 Apr 2009
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20 Aug 2009 11:30:06 IST
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u should hav been clear in asking the qst instead u should hav written the actual qst , and as fro the abv qst ans is B its true see here the cavity in the prblm may be considered as a superpostion of two balls one with +charge denisty and another -charge denisty , and then consider apoint P INSIDE A cavity and having R (r+)asa postion vector with respect to centre of ball and -R (r-) as a postion vector with respect to centre of cavity, thrn by prncpl of superpostion fnd the field inside cavity at a point p gvn by E=Er+Er-

New kid on the Block

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20 Aug 2009 22:40:15 IST
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ok...bt how - charge density comes..??there is no charge inside the cavity...does this mean that the guass's law has failed...??

Scorching goIITian

Joined: 24 Feb 2009
Posts: 214
23 Aug 2009 10:38:20 IST
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For this question use superimposibility of forces ...

Let O  and O' be the centres of the sphere and of the cavity respectively

Let a be the seperation bw the 2 centres

Lets take a point P inside the conductor at a distance x from O and y from O'

imagine that there was no cavity , then the field on P would be ....(1)

and due to the cavity individually it would be ....(2)

therefore the net electric field due to these two is    (1) - (2)

Using addition of vectors we get the above result as

which is a constant and is zero only if the 2 spheres are concentric which is not so in the given question and hence the answer is (b)

PLS VOTE

Cool goIITian

Joined: 17 Aug 2009
Posts: 30
26 Aug 2009 13:53:15 IST
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GOOD THINKING  Sachin , I too thought in the same way . Well, there is another interpretation for this problem . Just consider the gaussian surface containing the pt. P as described by Sachin. Calculate the field at that pt. due to the gaussian sphere. Let this be E. Now consider the cavity to contain a negative charge equal( in magnitude) to the charge contained by the portion of the sphere that was removed.Then find the field only due to the removed portion.Let this be E`. Then your required field (due to the remaining portion of the sphere) will be equal to E+E`, which u will find is constant.

This may seem ridiculus but it does serve ur purpose quite well. In case there was no charge in the sphere , but rather u were asked to find the gravitational field at that point then u would have to consider the mass of the cavity as negative. Had it been a conducting sphere then field at any pt. inside the cavity would have been 0.

Cool goIITian

Joined: 2 Sep 2012
Posts: 63
24 Apr 2013 22:50:49 IST
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if the confusion is about the applicability of gauss law,then gauss law is applicable only when:(i) electric field is either perpendicular or tangential allover in the gaussian surface (ii)magnitude of electric field on every point on gaussian surface is equal. in this case,it's not given that cavity is exactly at centre..so certainly when we form a gaussian surface around cavity ,the magnitude of electric field at all points of gaussian surface is not same,,so gauus law cannot be applied here.so a net electric field exists inside cavity (due to nonsymmetry around cavity) so electric field is uniform and nonzero ,.... no need to even solve it save your time..

Scorching goIITian

Joined: 10 Oct 2012
Posts: 237
25 Apr 2013 09:14:38 IST
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no! actually most you are confused over the main hint in the question, read carefully this question again. it is written "NON-CONDUCTING", ie it means that charge can reside inside the sphere too. (if it would have been conducting then charge only appear outside only), so here in this question, for any type of cavity created, there exist the electric field. (execpt in the cavity created symmetrically.)

Scorching goIITian

Joined: 10 Oct 2012
Posts: 237
25 Apr 2013 09:16:09 IST
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so, answer is non-zero and uniform.

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