Samagra G.

asked • 08/26/16

work performed on a charge

A solid non conducting hemisphere of radius R has a uniformly distributed positive charge of density t per unit volume . A negatively charge q is transferred from centre of its base to infinity . Dielectric constant of material of hemisphere is unity . Then prove that work performed in the process is (qtR2)/(4 ε)

1 Expert Answer

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Samagra G.

Can you tell me how you get the volume
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08/26/16

Samagra G.

You can show me that on diagram by uploading on imgur and giving the link . 
Can you tell also how can we find area of that small portion
 
Thank you
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08/26/16

Steven W.

tutor
Actually, since the spherical coordinate volume element is derived in several places on the internet, I think I can direct you to a site that has a diagram that is much better than I could (with my meager abilities) ever draw.

I am not able to post direct links here, but if you go to:

www -dot- vias -dot- org -slash- calculus -slash- img -slash- 12_multiple_integrals-418 -dot- gif

Replace the words "slash" with an actual slash, and "dot" with an actual dot. This is a diagram that shows how the volume element is derived in spherical coordinates. To get the volume of the hemisphere, you need to integrate one angle from 0 to 2π and the other from 0 to π.

And I made a couple errors that happened to work themselves out in the result. It should actually have been sinφ and not sinθ in the volume element, if φ represents the vertical angle, as I presented it. And the vertical angle is actually conventionally measured from the +z axis. But, even with these minor changes, the result will still be the same.

That is because you can choose which angle to integrate from 0 to 2π and which one from 0 to π, due to the spherical symmetry of the hemisphere in those directions.

See if that helps, and, if not, we can discuss it more. You can also search for "spherical coordinate volume element" on the internet to get more information. It is a standard calculus topic, and will show up in a lot of places. But I will be happy to help more, if needed.
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08/26/16

Shaad D.

On solving the integral for Vu'll get
V0= tR2/2£ not -tR2/4£ please check this out
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06/06/18

Steven W.

tutor
That is because I messed up the integral in two ways.  (1) It should be sin(φ), not sin(θ); and (2) the limits of integration on phi should be 0 to π/2, not 0 to π, since we are dealing with a hemisphere, not a sphere.

When I reworked with this, I still got 4 in the denominator, but the negative sign did vanish (as it should, since the charge density on the hemisphere is positive, leading to a positive V_o.  The negative work done by the electric field comes in because the sign of the point charge is negative.
 
I apologize for the confusion!
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06/06/18

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