Steven W. answered • 08/26/16
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Hi Samagra!
The most direct way to get at this problem, I think, is to realize that the work performed by an electric force on moving a charge from point to point is equal to qΔV, where q is the value of the charge being moved, and ΔV is the change in electric potential (in other words, voltage) between the two points.
When the charge is at infinity -- and considering this hemisphere of charge to be the only excess charge in space -- we can say the electric potential is zero (because the electric potential of this hemisphere goes to zero at infinity). So, the work done to move the charge from the center of the base to infinity is:
W = -q(Vf-Vo)= qVo
where Vo is the electric potential generated by the hemisphere at the center of its base. Now, we just have to calculate Vo.
Vo will be the sum of the contributions of every infinitesimal unit volume of charge at that position. I am going to use spherical coordinates to locate each infinitesimal volume unit on the sphere, with the center of the base of the hemisphere being the origin. In those terms, the volume of each unit is r2sinθdrdθdφ, where r is the radial distance to the charge, θ is its azimuthal angle (usually measured from the +x axis), and φ is its angle of elevation (usually measured from the xy plane). And each infinitesimal volume unit of charge is a distance r from the center of the base of the hemisphere.
The value of the charge in that infinitesimal volume unit is the product of the volume charge density t and the volume of that unit r2sinθdrdθdφ, so dq = t(r2sinθdrdθdφ).
Therefore, the contribution to the electric potential at the origin from a given unit volume of charge at (r,θ,φ) is:
dVo = kdq/d = kt(r2sinθdrdθdφ)/r = ktrsinθdrdθdφ
To get the total potential at the center of the hemisphere base, we have to integrate this dV expression over the entire hemisphere. This involves integrating the r direction from 0 to R, the θ direction for 0 to 2π and the φ direction from 0 to π (you could also switch the integration limits on θ and φ and get the same result). So, we have:
Vo = ∫0R ∫02π ∫0π ktrsinθdrdθdφ
k = the Coulomb constant, = 1/4πε (with ε = εo, if the dielectric constant is unity (=1))
I got Vo = -(t/4πε)(πR2) = -tR2/4ε
So, the work done by the electric field as the charge is moved from the base of the hemisphere to infinity is:
W = q(-tR2/4ε) = -qtR2/4ε
Note that this is the negative of what they want you to show. That is because this is the work done by the electric field. As the charge is moved from the center of the hemisphere base to infinity, there are two forces at work on it. The electric force of the hemisphere and the pulling force (actuating force) which is moving it away.
The work done by the electric force of the hemisphere on the charge is negative, as we showed above, This is because the hemisphere is positively charged, and the point charge q is negative. So, as the point charge is being pulled to infinity, the hemisphere is trying to pull it back. Therefore, the electric force of the hemisphere on the charge, as it is moved, is opposite the point charge's displacement. This makes for negative work, as we found.
Now, if we suppose the charge starts at rest at the center of the base of the hemisphere, and then ends at rest at infinity, the change in its kinetic energy overall must be zero (it starts and ends at rest). The work-energy theorem says that the net work done on an object must equal its change in kinetic energy. Hence, the work done by the two forces on the point charge as it is moved must add up to zero
Wpull+Welec = 0 --> Wpull = -Welec
So the work done by the pulling force must be equal in value and opposite in sign to the work done by the electric force during this move. So:
Wpull = qtR2/4ε
Hope this helps! If you have any questions about the process (and, without diagrams, I know it can be tricky), just let me know.
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.
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 V0 u'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.
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
Samagra G.
08/26/16