A Question of Electrostatic Repulsion
It’s been a while since I posted a question in the Cute Physics Problems folder so I thought I’d offer this one. It’s not particularly hard, but I think it’s quite instructive.
A thin spherical shell of radius r carrying a charge Q spread uniformly with constant surface density is split into two equal halves by a narrow planar cut passing through the centre as shown in the detailed diagram below:
Calculate the force arising from electrostatic repulsion between the two hemispherical shells, expressing your answer in terms of Q and r in SI units.
Answers through the Comments Box please. First correct answer wins a tomato*
*subject to availability
In the Dark 
March 9, 2023 at 12:20 am
The electric field at the surface of the sphere is
E = Q/r2,
directed radially.
Integrating F = ½ ∫Edq
over the uniform charge distribution across
the top half of the sphere gives a force
F = ⅛
Q/r2,
directed upwards.
Of the one eighth, one half is because the top has half the charge,
one half is the half in the expression, from removing the self-field
to obtain the externally applied field
(averaging the inner and outer fields),
and one half is from the geometric integral
(average of the vertical component).
The horizontal components sum to zero.
Add the usual 1/4πε0 to get SI units.
Or, maybe not, I haven’t been practicing lately.
March 9, 2023 at 5:21 am
Q2/r2, of course. And “up” is to the right.
March 9, 2023 at 8:53 am
I’m afraid that’s not the right answer…even accounting for the lack of SI units!
March 9, 2023 at 4:17 pm
Somehow the original post didn’t make it, and what was to be
a quick correction did.
Here it is again, complete with the error:
The electric field at the surface of the sphere is
E = Q/r², directed radially.
Integrating F = ½ ∫Edq
over the uniform charge distribution across
the top half of the sphere gives a force
F = ⅛ Q/r², directed upwards.
Of the one eighth, one half is because the top has half the charge,
one half is the half in the expression, from removing the self-field to
obtain the externally applied field (averaging the inner and outer fields),
and one half is from the geometric integral
(average of the vertical component).
The horizontal components sum to zero.
Add the usual 1/4πε0 to get SI units.
Or, maybe not, I haven’t been practicing lately.
So, the SI issue was addressed.
There is a YouTube video that does all integrals, but I couldn’t watch
the whole thing and I never see it get to an answer.
March 9, 2023 at 9:48 am
8/9 k Q/r2
with k the Coulomb constant
March 9, 2023 at 2:24 pm
Somehow the original post didn’t make it, and what was to be
a quick correction did.
Here it is again, complete with the error:
The electric field at the surface of the sphere is
E = Q/r²,
directed radially.
Integrating F = ½ ∫Edq
over the uniform charge distribution across
the top half of the sphere gives a force
F = ⅛ Q/r²,
directed upwards.
Of the one eighth, one half is because the top has half the charge,
one half is the half in the expression, from removing the self-field
to obtain the externally applied field
(averaging the inner and outer fields),
and one half is from the geometric integral
(average of the vertical component).
The horizontal components sum to zero.
Add the usual 1/4πε0 to get SI units.
Or, maybe not, I haven’t been practicing lately.
So, the SI issue was addressed.
There is a YouTube video that does all integrals, but I couldn’t watch
the whole thing and I never say it get to an answer.
March 9, 2023 at 4:11 pm
Somehow the original post didn’t make it, and what was to be a quick correction did.
Here it is again, complete with the error:
The electric field at the surface of the sphere is E = Q/r², directed radially.
Integrating F = ½ ∫Edq
over the uniform charge distribution across
the top half of the sphere gives a force
F = ⅛ Q/r², directed upwards.
Of the one eighth, one half is because the top has half the charge, one half is the half in the integral expression, from removing the self-field to obtain the externally applied field (averaging the inner and outer fields), and one half is from the geometric integral (average of the vertical component).
The horizontal components sum to zero.
Add the usual 1/4πε0 to get SI units.
Or, maybe not, I haven’t been practicing lately.
So, the SI issue was addressed.
There is a YouTube video that does all integrals, but I couldn’t watch the whole thing and I never saw it get to an answer.
March 11, 2023 at 9:25 am
Q^2 / (32pi epsilon_0 r^2)
March 11, 2023 at 10:43 am
This is the correct answer. Please send a stamped addressed envelope for the tomato.