Find the net force that the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere.

I am perfectly aware that this question has been asked many times (here), but I am surprised that I couldn't find a solution using integration by taking elements!

I tried to take elemental disks and then integrated it for the whole hemisphere, but my answer didn't match with the correct answer.

Here's my working:

It is known that for a uniformly charged disk, with charge density $\sigma$, the electric field at a point on it's axis is given by $E=\dfrac{\sigma}{2\epsilon_0}(1-\cos \theta)$, where $\theta$ is the angle between the axis and the line joining the point to the circumference of the disk.

And $\sigma= \dfrac{\mathrm{d}q}{\pi R^2 \sin^2 \theta}= \rho R ~\mathrm{d}\theta$ for an elemental disk, so the total field should be given by integrating the expression from $\theta=0$ to $\dfrac{\pi}2$ , so $$E=\int \mathrm{d}E= \dfrac{\rho R}{2 \epsilon_0} \int_0^{\frac{\pi}{2}} (1-\cos \theta) \mathrm{d} \theta \\ =\dfrac{\rho R}{2 \epsilon_0}\left(\dfrac{\pi}{2}-1\right) $$

Since $\rho=\dfrac{Q}{\frac{2}{3}\pi R^3}$, we get finally $E=\dfrac{3Q}{4\pi R^2 \epsilon_0}\left(\dfrac{\pi}{2}-1\right)$ and for the force we will multiply it by $Q$.

Which is totally different from the correct answer. What am I doing wrong ?

  • $\begingroup$ Your analysis finds the force assuming the upper hemisphere is a point charge at the origin, this approximation isn't valid. $\endgroup$
    – Triatticus
    Commented Jan 18, 2021 at 2:01

1 Answer 1


So finally I understand my mistake as pointed out by @Triatticus that I have assumed the northern hemisphere as a point charge, and that is not correct. It requires to calculate the force using integration and I am presenting a solution that doesn't require integration using polar coordinates as done in the linked post:

Proceeding by taking a gaussian shell of radius $r$, we apply the gauss law as $$ |E|4\pi r^2 = \dfrac{\rho\left(\dfrac{4}{3}\pi r^3\right)}{\epsilon_0} \implies E=\dfrac{\rho r}{3 \epsilon_0}$$ We will use $\rho=\dfrac{Q}{\frac{4}{3}\pi R^3}$ later on.

Now, we need the total force applied on each elemental charge $\mathrm dq$ due to this field. It is clear, that the force would be in the vertically up direction passing through the centre of the sphere due to symmetry.

Force applied on an elemental charge $\mathrm dq$ is given by $\mathrm d \overrightarrow{F}=\overrightarrow{E} \mathrm d q$ and so the total force is given by intgrating this expression, that is $$\overrightarrow{F}=\int \mathrm d \overrightarrow {F} =\int \overrightarrow E \mathrm dq=\dfrac{\rho}{3\epsilon_0}\int \overrightarrow{r} \mathrm d q$$

Now, what does the integral $\displaystyle \int \overrightarrow{r} \mathrm d q$ remind us of? Recall that in mechanics, the position of centre of mass of a body of mass $m$ is given by $\dfrac{\displaystyle \int \overrightarrow{r} \mathrm d m}{\displaystyle \int \mathrm d m}$ , so what we can infer from this is, using the similar analogy, we apply this whole force on the centre of mass of the northern hemisphere. It is a standard result that the centre of mass of a solid sphere is at the distance of $\dfrac{3R}{8}$ from the centre. So using $\displaystyle \int \overrightarrow{r} \mathrm d q=\dfrac{3QR}{16}$ , and plugging the value of $\rho$ in terms of $Q$, we get the net force as $$\boxed{F=\dfrac{3Q^2}{64 \pi \epsilon_0 R^2}}$$

  • $\begingroup$ Nice trick at the end, one point is that we are not using polar coordinates but spherical coordinates in linked pots. $\endgroup$
    – Babu
    Commented May 21, 2021 at 12:49

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