Two charged hemispheres

Find the force of repulsion between the two uniformly charged insulating hemi-spherical shells shown in the above figure.

Radius of larger shell = R R

Radius of smaller shell = r r

Q q s i n 2 α 16 π ϵ o R 2 \frac{Qqsin^{2}\alpha}{16\pi\epsilon_{o}R^{2}} Q q 8 π ϵ o R 2 \frac{Qq}{8\pi\epsilon_{o}R^{2}} Q q c o s 2 α 16 π ϵ o R 2 \frac{Qqcos^{2}\alpha}{16\pi\epsilon_{o}R^{2}} Q q 16 π ϵ o R 2 \frac{Qq}{16\pi\epsilon_{o}R^{2}}

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1 solution

Mark Hennings
May 8, 2019

Call the larger hemispherical shell H H and the smaller shell K K . Imagine a second uniformly charged large hemispherical shell H H' which, together with H H , forms a complete uniformly charged spherical shell with total charge 2 Q 2Q . Let F H K \mathbf{F}_{HK} be the electrostatic force between H H and K K , and let F H K \mathbf{F}_{H'K} be the electrostatic force between H H' and K K . Since the electric field inside a hollow uniformly charged hemisphere is zero, we deduce that F H K + F H K = 0 \mathbf{F}_{HK} + \mathbf{F}_{H'K} = \mathbf{0} .

If we now imagine a second uniformly charged small hemisphere K K' , which, together with K K , forms a complete uniformly charged spherical shell of charge 2 q 2q , and let F H K \mathbf{F}_{HK'} be the electrostatic force between H H and K K' then, by symmetry, F H K = F H K = F H K \mathbf{F}_{HK'} = -\mathbf{F}_{H'K} = \mathbf{F}_{HK} . Thus F H K = 1 2 ( F H K + F H K ) \mathbf{F}_{HK} = \tfrac12\big(\mathbf{F}_{HK} + \mathbf{F}_{HK'}\big) , and this last is the electrostatic force between H H and the entire smaller spherical sphere. Since the electric field of a uniformly charged spherical shell is, outside that shell, the same as the electric field of a point charge (of the same charge as that on the shell) at the centre of the shell, we deduce that F H K \mathbf{F}_{HK} is equal to 1 2 G 2 q = G q \tfrac12\mathbf{G}_{2q} = \mathbf{G}_q , where G u \mathbf{G}_u is the electrostatic force between H H and a point charge u u at the centre of the hemispherical shell H H .

It is standard bookwork therefore that F H K = G q = Q q 8 π ε 0 R 2 \big|\mathbf{F}_{HK}\big| \; = \; |\mathbf{G}_q| \; = \; \boxed{\frac{Qq}{8\pi\varepsilon_0R^2}} with the force acting along the line of symmetry of the larger hemispherical shell.

0.5G[2q]=(0.5)(2q)(Q)/4πεr^2=qQ/4πεr^2 ??

wing yan yau - 1 year, 6 months ago

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G q |\mathrm{G}_q| is the force on a charge q q from a hemisphere only.

Mark Hennings - 1 year, 6 months ago

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but G[q]=0.5G[2q] right?

wing yan yau - 1 year, 6 months ago

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@Wing Yan Yau Yes, but the force between a hemisphere carrying a charge Q Q at the point charge q q (at the centre of the hemisphere) is G q = 0 1 2 π cos θ Q 2 π R 2 q 4 π ε 0 R 2 × 2 π R sin θ × R d θ = Q q 8 π ε 0 R 2 |\mathbf{G}_q| = \int_0^{\frac12\pi} \cos\theta \frac{\frac{Q}{2\pi R^2}q}{4\pi \varepsilon_0 R^2} \times 2\pi R \sin\theta \times R\,d\theta \; =\; \frac{Qq}{8\pi \varepsilon_0 R^2} acting along the axis of the hemisphere.

Mark Hennings - 1 year, 6 months ago

Same solution as provided in the Puzzling problem book

raj abhinav - 1 year ago

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If you say so...

Mark Hennings - 1 year ago

Isn't there any other way to do it @Mark Hennings

raj abhinav - 1 year ago

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Well obviously, the force could be expressed as a multiple integral over two hemispheres, but I don't fancy that; do you?

Mark Hennings - 1 year ago

actually i am not acquainted with such mathematics right now

raj abhinav - 1 year ago

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