# Force between charged hemispherical shells

Consider two hollow uniformly charged hemispherical shells forming a complete sphere. The problem is to calculate the force of repulsion between these two shells. A common method is using electrostatic pressure which we can calculate to be equal to $$\frac{\sigma^2}{2\epsilon_{0}}$$ for a small surface elemental. Now as we do the integral of this over the curved surface of a hemisphere and add up the forces vectorially we calculate that it turns out the same as multiplying by the projected area (= $$\pi r^2$$). So the force is $$\frac{\sigma^2}{2\epsilon_{0}} \pi r^2$$ . But while doing so havent we included the effect of pressure produced by the charge on the hemispherical shell on itself? Isn’t the electrostatic pressure produced by the entire remaining region of the sphere except the small elemental we took, which includes the force due to the charges on both hemispheres. Also ,due to these reasons if this is a wrong way to approach the problem ; how would we proceed in finding the force?

It turns out that the self force is zero in electrostatics (also true for magnetostatics). You can prove it using the Maxwell stress tensor (generalisation of electrostatic pressure): $$f = \nabla\cdot \sigma \\ f = \rho E \\ \sigma = \epsilon_0 \left(E\otimes E-\frac{1}{2}E^2 \mathbb 1\right)$$ with $$E$$ the static field generated by the charge density $$\rho$$, i.e. following: $$\nabla\cdot E = \rho/\epsilon_0\\ \nabla\times E = 0$$ In integrating of all of space, this transforms your force into a boundary term. Since your charge is localised, $$E=O(r^{-2})$$ so $$\sigma = O(r^{-4})$$ so even if the surface at infinity scales as $$O(r^2)$$, the boundary term goes to zero.
• The first part of the answer is a general treatment, not just for your hemisphere. You can integrate only on the support of the charge density (the hemisphere in your case), but it'll be harder to argue why the self force is zero. Note that the force balance is valid everywhere. Furthermore, integrating the LHS beyond the support still gives you the same result as $\rho=0$ in the extra region. Thus, integrating the RHS over all space also gives you the total self-force.