Electric field due to point charge enclosed by a hollow conducting sphere?

Discussion in 'Homework Help' started by Star Wars, Aug 2, 2013.

  1. Star Wars

    Thread Starter New Member

    Oct 5, 2012
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    The electric field due to point charge +Q at a distant point P is V volt/meter.
    Now the point charge +Q is enclosed by the hollow conducting sphere,

    a. What is the charge in the inner and outer surface of the enclosing sphere?
    b. What is the flux inside and outside of the enclosing sphere?
    c. What is the field at the point P (0 or V) and why ?
     
  2. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    Ok...I give up. What is it?
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    This belongs in the Homework Help forum. You will get more attention there. I'll as a mod to move it.

    In general, you need to make an honest attempt to work out your own problems. We will then use that as a springboard to help guide your efforts and correct any misconceptions you are making.
     
  4. bertus

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  5. Star Wars

    Thread Starter New Member

    Oct 5, 2012
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    Sorry, I intended to post this Q. in Physics section.
     
  6. WBahn

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    Mar 31, 2012
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    The Physics section is not intended for homework problems, but rather non-homework physics discussions. You query is better served in the Homework section because the number of people that read and respond to the Physics section is quite small in comparison.

    But in either place, you will need to put in some thought and effort into trying to work toward a solution yourself and showing what you have done.
     
  7. Star Wars

    Thread Starter New Member

    Oct 5, 2012
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    Ok, I think... As shown in attached figure here..

    a) The charge in the inner and outer surface of the enclosing hollow conducting sphere will be as shown in the figure - inner (-Q) outer(+Q).

    b) The net flux inside the conducting hollow sphere is zero due to +Q point charge and -Q(on the inner surface of hollow sphere).
    And the net flux outside the conducting hollow sphere is +Q (in any Guassian surface of course) due to the +Q charge induced on the outer surface of hollow sphere.

    c) The electric field on the point P is same i.e. V volts/meter due to the +Q charge on the outer surface of the hollow conducting sphere.

    What do you say?
     
    Last edited: Aug 3, 2013
  8. studiot

    AAC Fanatic!

    Nov 9, 2007
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    You should clearly distinguish between the (lines of) flux and the charge. The charge is not moving.
     
  9. studiot

    AAC Fanatic!

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    Well, you are going in the right direction, but why have you noted flux in charge units in (b)?
     
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  10. Star Wars

    Thread Starter New Member

    Oct 5, 2012
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    Thanks for response,

    In part b) I have calculated the flux through the red Guassian surface, that's why the total flux through this Guassian surface is the net charge enclosed by it (according to Guass's theorem).

    Is it OK?
     
  11. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Your reasoning towards the answers is OK.

    However your statement of Gauss' Law is not, because you have the units wrong.


    I don't know in what order you are being taught about electric fields.

    We have E, D, F, V as vector fields to choose from to calculate a 'flux', but none of the definitions of flux are in coulombs.


    Gauss' Flux Theorem states

    \smallint {\smallint _{cs}}E\cos \theta dS = \sum\limits_i {\frac{{{q_i}}}{{{\varepsilon _0}}}}

    That is the surface integral of the normal component of field E, taken over a closed surface, equals the sum of the charges enclosed within that volume divided by {{\varepsilon _0}}

    Thus the units of the flux are not coulombs as you have stated in (b)

    What are the units of {{\varepsilon _0}} ?

    The units of flux are ?

    Look here to read about the units.
    http://en.wikipedia.org/wiki/Electric_flux
     
    Last edited: Aug 4, 2013
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  12. WBahn

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    Please take what studiot is saying about the units issue seriously. Meanwhile, I'm going to sit back and enjoy someone else harping on units for a change. :p

    One nitpick on your diagram. It is drawn as though there is +1Q at the center, -6Q on the inner surface of the shell, and +6Q on the outer surface. I know what you are trying to say, but good engineering is as much about good and clear communication as anything else and diagrams can make things either very clear or they can result in very misleading impressions. I would recommend putting a single -Q just inside the shell and drawing an arrow from the text to the inner side of the shell. Then do the same for a single +Q located just outside the shell with an arrow from it to the outer surface. This will pretty clearly indicate that you are talking about a total charge that is, by symmetry as well as implication, evenly distributed over the surface pointed to.
     
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  13. Star Wars

    Thread Starter New Member

    Oct 5, 2012
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    Thanks, I need to be careful for units.
     
  14. Star Wars

    Thread Starter New Member

    Oct 5, 2012
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    Sure.

    I apologize for this.
     
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