# Why did the author choose this as the electric field?

In the image below, the author works out an example. However, I don't understand how he arrived at $$\vec{E}$$. In the problem, it says that the charge is uniformly distributed, but somehow he says that the distance from the source point to field point everywhere on the northern hemisphere is at a distance $$R$$ away. My guess might be that he is using this concept (correct me if I say it wrong): when you have a solid sphere that is uniformly charged, you can pretend that the entirety of the charge is located at its center. Then, you can treat this as a discrete charge which allows you to use a fixed distance to any field point.

However, if this is actually the concept he used I have a problem with it. The southern hemisphere is also charged and this side of the sphere will have a different value for its contribution to the $$\vec{E}$$ field. In fact, every point on the sphere will have an effect on every other point on the sphere. I know that we are dealing with a continuous distribution of charge here but to help me better visualize the charge I think of the scenario as a HUGE (see: uncountable) collection of point charges. Now I ask the question: How does this point charge affect that point charge? From this perspective, every point on the sphere affects the northern hemisphere.

Could someone help me understand this?

For clarification: I'm not asking about the final solution, I'm only asking about one step in his solution. I don't care about the net force - I only want to know how he arrived at $$\vec{E}$$.

• The reference textbook is "Introduction to Electrodynamics" David J. Griffiths, 4th Edition. It's not a problem but an Example of how to use the Maxwell stress tensor and Eq. 8.21 to determine net forces : $$\mathbf{F}=\oint_S \overleftrightarrow{\mathbf{T}}\boldsymbol{\cdot} d \mathbf{a} \qquad \textrm{(static).} \tag{8.21}$$ So, read carefully 'CHAPTER 8 Conservation Laws' and the solution given in the Example. Commented Jan 24, 2017 at 22:29
• I downvoted your question because of no effort by your side. Commented Jan 24, 2017 at 22:38
• Between others by "no effort" I mean that you don't even turn from page 364 to pages 365,366 to read the whole solution. It seems that you have a photocopy of page 364 only and you think that the answer is the red circled (by you) equation. The answer is on page 365 $$F=\dfrac{1}{4\pi \epsilon_{0}}\dfrac{3Q^{2}}{16R^{2}} \tag{8.26}$$ identical to that given also in problem 2.47 (page 108), as you could ascertain if you would have the textbook on hand in the future. Commented Jan 24, 2017 at 23:13
• Again, I repeat the comment I made to the given answer below. I'm not asking about how the author found the net force, I'm asking about 1 simple step in his solution. I don't see why both of you seem to think I am asking about the final answer, I'm asking about step 2 of his reasoning. Commented Jan 24, 2017 at 23:15
• I take back the downvoting. Commented Jan 25, 2017 at 0:14

## 3 Answers

You are not aswering the question that was asked. You were not asked to compute the force due to the electric field by adding up the force on every bit of the upper hemisphere due to the charges in the lower half. You were instead asked to use the Maxwell stress tensor. The magic of the stress tensor is that you only need to know the total electric field (due to the charges in both hemispheres) on the $surface$ of the part of the object of interest. The total electric field on the curved surface of the upper is the radially outward field of the entire sphere of charge. (you will need to do some work to compute the E field on the flat surface). The Maxwell-stress answer will, of course, be same as the force due to the charges in the lower hemisphere on those in the upper hemisphere. This because the total force on the charges in the upper hemisphere due to the charges only in the upper hemisphere is zero.

• Note that I wasn't asking about help with the entirety of the problem, only the step where he finds $\vec{E}$. Commented Jan 24, 2017 at 22:57
• Then it is trivial. His answer is the electric field due to the entire solid sphere of charge, which, at its surface is equal to the that of a point charge at its centre. Commented Jan 25, 2017 at 13:38

To respond to a previous comment asking where the southern hemisphere is: the graphic in Griffiths is misleading. Note that the problem states to find the net force on the northern hemisphere of a charged solid sphere.

That's why the electric field was taken to be $$\vec{E}=\frac{1}{4\pi\epsilon_0}\cdot\frac{Q}{R^2}$$. If one only has a charged hemisphere, it is by no means obvious what the electric field will be on the surface. In fact, it likely isn't even constant on the surface due to a lack of spherical symmetry.

My guess might be that he is using this concept (correct me if I say it wrong): when you have a solid sphere that is uniformly charged, you can pretend that the entirety of the charge is located at its center.

No, that's not the concept. The concept is that physical laws are isotropic: any asymmetry in the result of a configuration must result from asymmetry of that configuration. Since the charge is spherically symmetric, the electric field must be spherically symmetric. If there were any point on the surface of the sphere that has a larger electric field than another point, then those points would be somehow distinguished, but there is nothing in the problem statement that allows these points to be distinguished. Thus, each point on the surface must have the same magnitude of electric field. We can calculate this magnitude by treating the sphere as being a Gaussian surface.