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

2 votes
349 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.

asked Jan 28
edited Feb 5
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## 2 Answers

4 votes

Best answer The problem asks to calculate the force on one part due to other. But the electric field I'm using to do this calculation is the result E of complete solid non-conducting sphere. So why are we considering E of the northern hemisphere to calculate force it experience from the southern hemisphere?

This is a very good question.

One way of finding the total electrostatic force on the N hemisphere is to calculate the vector force $F_{ij}=k q_i q_j / r_{ij}^2$ on every charge $q_i$ in the N hemisphere due to every charge $q_j$ in the S hemisphere, then take the vector sum of all forces $F_{ij}$ :

$$F_N=\sum_{i \in N} \sum_{j \in S} F_{ij}$$

This double sum or integral is very difficult to calculate, for two reasons :

1. the general expression for the force $F_{ij}$ between every pair of 2 charges is a complicated function using the Cartesian co-ordinate system. even more so when using spherical co-ordinates ;

2. each of the 2 charges $q_i, q_j$ in every pair has 3 co-ordinates, so in total there will be a sum ranging over 6 possible co-ordinates.

An alternative method of solution (used in Jorge Daniel's answer) is to sum the total electrical force $F_i$ on each charge $q_i$ in the N hemisphere due to all of the charges in both the S and N hemispheres :

$$F_N=\sum_{i \in N} F_i=\sum_{i \in N} \sum_{j \in N,S} F_{ij}$$

Whereas $F_{ij}$ is a very complicated expression, the total force $F_i$ on each charge is very much simpler because it is known to be radial and depends only on the radial distance of the charge $q_i$ from the centre of the sphere.

As you rightly point out, this means that within the expression for $F_i$ we will include the force $F_{ij}$ on charge $q_i$ due to other charges $q_j$ which are also in the N hemisphere.

However, if charge $q_j$ is also within the N hemisphere ($j \in N$), then when we take the sum over all charges $q_i$ in the N hemisphere ($i \in N$), this sum will include not only $F_{ij}$ but also the equal and opposite force $F_{ji}$ acting on $q_j$ due to charge $q_i$. These 2 contributions to the total force on all particles in the N hemisphere will cancel out because $F_{ij}=-F_{ji}$.

On the other hand, if charge $q_j$ is in the S hemisphere ($j \in S$) then the force $F_{ji}$ will not be included in the first summation over $i \in N$.

To make the above explanation clearer, suppose we have a system of only 4 charges $q_i$ with $i=1 \to 4$. Between every pair of charges there is an electrical force $F_{ij}$ meaning the force that charge $j$ exerts on charge $i$.

Suppose that charges $i=1 , 2$ constitute one object (which we call "the N hemisphere") and charges $i=3, 4$ constitute a second object ("the S hemisphere"). Then the total force on the N hemisphere due to the S hemisphere is $$F_N=F_{13}+F_{14}+F_{23}+F_{24}$$ The total forces on charges $q_1, q_2$ due to all other charges in both hemispheres (ie summing over $j \in N, S$) are $$F_1=F_{12}+F_{13}+F_{14}$$ $$F_2=F_{21}+F_{23}+F_{24}$$ If we add the forces on charges $q_1$ and $q_2$ (summing over $i \in N$) we get $$F_N'=F_1+F_2=(F_{12}+F_{21})+F_{13}+F_{14}+F_{23}+F_{24}=F_N$$ because $F_{12}=-F_{21}$.

Thus both methods give the same answer.

answered Mar 4 by (28,746 points)
selected Mar 10 by n3
1 vote

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 (666 points)