# How is Newton’s third law valid in the case of a charged sphere enclosed within another charged sphere?

Let's say we have two point charges q1 and q2 both of which are positive, according to Newton's third law the force one exerts on the othe is equal to the force the other exerts on the first. Therefore

$$F_1 = q_1 E_2$$

$$F_2 = q_2 E_1$$

$$F_1= F_2$$

Therefore:

$$q_1 E_2 = q_2 E_1$$

Now let's say that we have two concentric spheres both of which are infinitely thin and have a positive charge:

Let's say that the inner sphere is Sphere 1 and the outer sphere is Sphere 2

According to Gauss's law The Electric field due to sphere 2 is zero everywhere inside sphere 2. The electric field due to sphere 1 is zero inside sphere 1 however it is not zero outside of it.

The force of the charges from sphere 1 acting on sphere 2 is:

$$E_{sphere1} q_{sphere2}$$

while the force of the charges from sphere 2 acting on sphere 1 is:

$$E_{sphere2} q_{sphere1}$$

However the electric field from sphere 2 is zero everywhere inside itself.

If the electric field due to sphere 1 is non-zero, and the electric field does push charges on the second sphere outward, how is Newton's third law obeyed if the outer sphere doesn't push back?

how is newton's third law obayed if the outer sphere doesn't push back?

Newton’s 3rd law is obeyed in many different but equivalent ways.

First, if we consider the spheres as a whole the net force on the outer sphere is zero since the force is constant and radially outward and the force on the inner sphere is also zero. Clearly $$0=-0$$. So overall, as an entire object Newton’s 3rd law holds.

Now, more interesting is to consider each individual charge. For any individual charge on the inner sphere and any individual charge on the outer sphere the force between them is given by Coulomb’s law. By Coulomb’s law the magnitude of the force is the same on each and the direction is opposite. So individually, for each pair of charges Newton’s 3rd law holds.

Finally, consider one charge on the inner sphere and the entire outer sphere. The net force on the inner charge is zero by Newton’s shell theorem, and it turns out that the net force on the shell is also zero. The part of the outer shell that the inner charge is closest to has the strongest force, but there is more charge further away. Over the entire shell the force again integrates to 0. So for every possible combination of charges, Newton’s 3rd law holds.

Now, what you may be seeing is that the stress on the outer sphere due to the inner sphere is non-zero, while the stress on the inner sphere due to the outer sphere is zero. However, Newton’s 3rd law says nothing about stresses. It only asserts that the forces are equal and opposite. The stresses are unequal, and that is perfectly consistent with Newton’s 3rd law

• Let's say that the outer sphere is a balloon and it is "let go" would it grow in size? (until it is balanced by the restoring elastic force of course). If the balloon does grow doesn't that mean that the inner sphere is excering an "outward" force on each point of the outer sphere causing it to grow? However if the inner sphere is a balloon, the outer spher wouldn't cause the inner balloon to contract because of gauses law Commented Sep 17, 2020 at 5:31
• @Sasha1296 I discussed exactly this point in the last paragraph of the answer. Yes, the outer sphere would grow. Yes, the growth is caused by stress from the inner sphere. Yes, the inner sphere does not experience any stress and thus it will not shrink. All of that is consistent with Newton’s 3rd law. Notice that as the outer sphere grows, its center of mass is unchanged. This demonstrates that the force on the outer shell is 0. The force is 0 as required by Newton’s 3rd law, the stress is not 0 which is OK by Newton’s 3rd
– Dale
Commented Sep 17, 2020 at 11:03
• ok, I guess that makes sense, I guess it's just weird that stress isn't conserved Commented Sep 17, 2020 at 22:10

You are simplifying things too much here. To calculate the force sphere 1 exerts on sphere 2 you can't just do $$F=qE$$ because we are dealing with vectors here. In general, you will need to do an integral (sum) of all of the vector forces over sphere 2 due to all of the charges on sphere 1, and you will find that everything cancels out for the same reason that there is no net field from sphere 2 inside of sphere 2. Newton's laws still hold here.

If the spheres are concentric then you can also use a symmetry argument. The field will be constant in magnitude across sphere 2, and each point on sphere 2 will have a point on the opposite side of the sphere where the field points in the opposite direction. Therefore, everything cancels out.

Remember, electric fields follow the law of linear superposition, so since Coulomb's law for point charges follows Newton's third law, we are always guaranteed that it will also hold for electrostatic interactions of systems of many charges as well.

• As another comment I've made, let's say case 1) the outer spher is now a balloon that is free to expand, wouldn't the electric field from the inner sphere cause it to expand? However if only the inner sphere was the ballon, the electric field from the outer sphere wouldn't cause it to contract Commented Sep 17, 2020 at 5:35
• Am I overthinking this? I guess the outer balloon doesn't gain any net momentum as the total force summed over all directions is zero, but it still seems weird that an inner sphere can cause the outer sphere to expand but the outer sphere doesn't "push back" and cause the inner sphere to contract Commented Sep 17, 2020 at 5:41