A metal sphere of radius $R $ carries a total charge $Q $. What is the force of repulsion between the "northern" hemisphere and the "southern" hemisphere?
An important keyword here is "A metal sphere," meaning that we are dealing with a conductor. As such, it follows that
\begin{align*} \mathbf{E}_{\text{Inside} } &= \mathbf{0} \hat{\mathbf{r}}\\ \mathbf{E}_{\text{Outside} } &= \frac{1}{4\pi \epsilon _0}\frac{Q}{r^2 }\hat{\mathbf{r}} \end{align*}Where the first comes from the fact that it is a conductor, and ideal conductors (see page 95) are such that $\mathbf{E}=\mathbf{0} $ inside (property 1 of conductors). Furthermore, since conductors hide the internal charge configuration, we can simply treat this sphere as a uniformly charged sphere. The field of such a sphere is the one given above. (For a derivation of this result, see Example 2.3, page 69.)
Recall the force per unit area equation:
\[\begin{split} \mathbf{f} &= \sigma \mathbf{E}_{\text{avg} } \\ &= \frac{1}{2}\sigma (\mathbf{E}_{\text{above} }+\mathbf{E}_{\text{below} }) \end{split} \tag{Eq. 2.50, page 100}\]$\sigma $ will be:
\[\begin{split} \sigma &= \frac{\text{Total charge inside conductor} }{\text{Conductor's surface area} } \\ &= \frac{Q}{4\pi R^2 } \end{split} \](A result that's familiar to us after doing problem 2.40!) And because $\mathbf{E}_{\text{below} }=\mathbf{0} $, we see that:
\[\begin{split} \mathbf{f} &= \sigma \frac{1}{2}\mathbf{E}_{\text{above} } \\ &= \frac{1}{2}\frac{Q}{4\pi R^2 } \mathbf{E}_{\text{Outside} }\Big|_{r=R} \\ &= \frac{1}{2}\frac{Q}{4\pi R^2 }\frac{1}{4\pi \epsilon _0}\frac{Q}{R^2 }\hat{\mathbf{r}} \end{split} \]To get the force of repulsion between the "northern" hemisphere and the "southern" hemisphere, we need only the vertical component of the force per unit unit area:
\[f_z=|\mathbf{f}|\cos \left( \theta \right) \]We finish by integrating this over the northern hemisphere:
\[\begin{split} F_z &= \int f_zda \\ &= \int_{\phi =0}^{2\pi } \int_{\theta =0}^{\pi /2} \left[ \frac{1}{2}\frac{Q}{4\pi R^2 }\frac{1}{4\pi \epsilon _0}\frac{Q}{R^2 } \right]\cos \left( \theta \right) \left( R^2 \sin \left( \theta \right) d\theta d\phi \right) \\ &= \frac{1}{2}\frac{Q}{4\pi R^2 }\frac{1}{4\pi \epsilon _0}\frac{Q}{R^2 }R^2 (2\pi )\int_{\theta =0}^{\pi /2} \cos \left( \theta \right) \sin \left( \theta \right) d\theta \\ &= \frac{Q }{4\pi \epsilon _0}\frac{Q}{4R^2 } \left( \frac{1}{2}u ^2 \right) _0^{1} \\ &= \frac{Q^2}{32\pi R^2 \epsilon _0} \end{split} \]Where $u $-substitution was used as follows: Let $u=\sin \left( \theta \right) $, then $du =\cos \left( \theta \right) d\theta $, so $d\theta =\frac{du}{\cos \left( \theta \right) } $ and the integral simplifies to $\int_{0}^{\pi /2} \sin \left( \theta \right) \cos \left( \theta \right) d\theta = \int_{0}^{1} du$.
We see than that the force of repulsion between the "northern" hemisphere and the "southern" hemisphere is
\[F_z=\frac{Q^2}{32\pi R^2 \epsilon _0} \]