# Force that one half of uniformly charged solid sphere exerts on other half - Griffith 2.43

34 views

I got the result mentioned, by considering the electric field as $\dfrac{\rho\cdot \vec{r}}{3\epsilon_o}$, and integrating carefully.

As problem asks to calculate the force on one part due to other, but the electric field I'm using here is the result of complete solid non conducting sphere, so why are we considering $E$ of the northern hemisphere to calculate force it experience.

edited Feb 5

If I am not mistaken, you are wondering why we use $\vec{E}$ so as to compute the net force that the southern hemisphere exerts on the northern hemisphere.

A physical interpretation of $\vec{E}$ would be the force per unit charge exerted on a test charge. In this exercise we deal with charge distributions; $\vec{E}$ can be interpreted as the force per unit volume per charge distribution.

This electric field is triggered by the charge enclosed in the sphere.

Using the volume charge density in the R sphere:

$$\rho = \frac{Q}{V} = \frac{3Q}{4 \pi R^3}$$

This problem presents spherical symmetry and allows us to set up a Gaussian (smaller; r < R)
sphere and compute the electric field straightforwardly. Thus:

$$q = \rho v = \frac{Qr^3}{R^3}$$

Where $v$ is the Gaussian spherical volume and the charge q is the charge enclosed in the (smaller; r < R) spherical Gaussian surface. Now by Gauss' Law:

$$EA = \frac{q}{\epsilon}$$

$$E = \frac{q}{A \epsilon} = \frac{Q r}{4 \pi \epsilon R^3}$$

Now it is just about using the force per unit volume definition:

$$f = \rho E = \frac{3r}{\epsilon}(\frac{Q}{4 \pi \epsilon R^3})^2$$

Now it is useful to set an infinitesimal volume element:

$$d\tau = r^2 sin dr d\theta d\phi$$

The net force is just the contribution of the force in the z direction

$$f_z = f cos \theta \hat{z}$$

So the net force in the z direction on the infinitesimal volume element is:

$$f_z = fcos \theta d\tau$$

And finally to get the total net force exerted on the northern hemisphere you just have to integrate:

$$F = \int_0^R \int_0^{\frac{\pi}{2}} \int_0^{2\pi} f cos \theta d\tau = \frac{3Q^2}{64 \pi \epsilon R^2}$$

answered Feb 3 by (528 points)