Steven W. answered • 08/26/16

Physics Ph.D., *professional*, easygoing, 6000+ hours tutoring physics

_{f}-V

_{o})= qV

_{o}

_{o}is the electric potential generated by the hemisphere at the center of its base. Now, we just have to calculate V

_{o}.

_{o}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 r

^{2}sinθdrdθdφ, 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.

^{2}sinθdrdθdφ, so dq = t(r

^{2}sinθdrdθdφ).

_{o}= kdq/d = kt(r

^{2}sinθdrdθdφ)/r = ktrsinθdrdθdφ

_{o}= ∫

_{0}

^{R}∫

_{0}

^{2π}∫

_{0}

^{π }ktrsinθdrdθdφ

_{o}, if the dielectric constant is unity (=1))

_{o}= -(t/4πε)(πR

^{2}) = -tR

^{2}/4ε

^{2}/4ε) = -qtR

^{2}/4ε

_{pull}+W

_{elec}= 0 --> W

_{pull}= -W

_{elec}

_{pull}= qtR

^{2}/4ε

Samagra G.

08/26/16

Steven W.

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.

08/26/16

Shaad D.

_{0 }u'll get

_{0}= tR

^{2}/2£ not -tR

^{2}/4£ please check this out

06/06/18

Steven W.

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.

06/06/18

Samagra G.

08/26/16