Question 8.2: Determine the net force on the “northern” hemisphere of a un...

F=\oint_{S}^{}{\overleftrightarrow{T}.da }      (static).     (8.21)

The boundary surface consists of two parts—a hemispherical “bowl” at radius R, and a circular disk at θ = π/2 (Fig. 8.4). For the bowl,

and

In Cartesian components,

T_{zx}=\epsilon_0E_zE_x=\epsilon _0\left(\frac{Q}{4\pi\epsilon _0R^2} \right)^2 \sin\theta\cos\theta\cos\phi,\\T_{zy}=\epsilon_0E_zE_y=\epsilon _0\left(\frac{Q}{4\pi\epsilon _0R^2} \right)^2 \sin\theta\cos\theta\cos\phi,\\T_{zz}=\frac{\epsilon _0}{2}(E^2_z-E^2_x-E^2_y) =\frac{\epsilon _0}{2}\left(\frac{Q}{4\pi\epsilon _0R^2} \right)^2 (\cos^2\theta-\sin^2\theta).            (8.22)

The net force is obviously in the z-direction, so it suffices to calculate

The force on the “bowl” is therefore

F_{bowl}=\frac{\epsilon _0}{2}\left(\frac{Q}{4\pi\epsilon _0R} \right)^2 2\pi \int_{0}^{\pi/2}{\sin\theta\cos\theta d \theta}=\frac{1}{4\pi\epsilon _0}\frac{Q^2}{8R^2}.         (8.23)

Meanwhile, for the equatorial disk,

da=-rdrd\phi\hat{z},    (8.24)

and (since we are now inside the sphere)

Thus

and hence

The force on the disk is therefore

F_{disk}=\frac{\epsilon _0}{2}\left(\frac{Q}{4\pi\epsilon _0R^3} \right)^2 2\pi \int_{0}^{R}{r^3dr}=\frac{1}{4\pi\epsilon _0}\frac{Q^2}{16R^2}.      (5.25)

Combining Eqs. 8.23 and 8.25, I conclude that the net force on the northern hemisphere is

F=\frac{1}{4\pi\epsilon _0}\frac{3Q^2}{16R^2}.       (8.26)

Incidentally, in applying Eq. 8.21, any volume that encloses all of the charge in question (and no other charge) will do the job. For example, in the present case we could use the whole region z > 0. In that case the boundary surface consists of the entire xy plane (plus a hemisphere at r =∞, but E = 0 out there, so it contributes nothing). In place of the “bowl,” we now have the outer portion of the plane (r > R). Here

(Eq. 8.22 with θ = π/2 and R →r ), and da is given by Eq. 8.24, so

and the contribution from the plane for r > R is

the same as for the bowl (Eq. 8.23).