Force between charged Ring and Rod, disproved Newtons 3rd law?

In summary, the electric field due to the ring is:The force on the rod due to this Electric field produced by the ring is:Calculating the field due to the rod:The net electric field in the plane of the ring will have both a x and y component because the rod is not infinitely long in both direction, the bottom of the rod is at the center of the ring and the other end tends to infinity(in the +y direction).
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Homework Statement
A system consists of a thin charged wire ring of radius ##R## and a very long uniformly charged rod oriented along the axis of the ring, with one of its coinciding with the centre of the ring. the total charge of the ring is equal to ##q##. The charge of the rod(per unit length) is equal to ##\lambda##. Find the interaction force between the ring and the rod
Relevant Equations
##\vec F = q\vec E##
This is the initial setup of the problem:
1638026276093.png

The electric field due to the ring is:
$$E = \int\frac{k(dq)}{(\sqrt{R^2 + x^2})^2}\frac{x}{\sqrt{R^2 + x^2}} = \frac{kqx}{(R^2 + x^2)^{3/2}}$$
the force on the rod due to this Electric field produced by the ring is:
Consider a differential element ##dq## of the rod on which the force is:
1638026372669.png

$$dF = E(dq) = E\lambda(dx)\frac{kqx\lambda(dx)}{(x^2 + R^2)^{3/2}}$$
The total force on the rod by the ring will be:
$$F = kq\lambda\int_{0}^{\infty}\frac{x}{(x^2 + R^2)^{3/2}} dx = -kq\lambda\left[\frac{1}{\sqrt{x^2 + R^2}}\right]_{0}^{\infty}=\frac{kq\lambda}{R} = \frac{q\lambda}{4\pi\epsilon_0 R}$$
this is the correct answer given in the book. But if we calculate the field due to rod on the ring and then calculate the Force acting on the ring, by Newtons 3rd law we should get the same force

calculating the field due to the rod:
1638026450012.png

$$|d\vec E| = \frac{k(dq)}{(x^2+R^2)}$$
$$d\vec E = dE_x\hat i + dE_y\hat j$$
$$dE_x = dEcos\theta = \frac{k(dq)}{(x^2 + R^2)}\frac{R}{\sqrt{R^2 + x^2}} = \frac{k\lambda(dx)R}{(R^2+x^2)^{3/2}}$$
$$E_x = k\lambda R\int_{0}^{\infty}\frac{dx}{(R^2 + x^2)^{3/2}}=\frac{k\lambda}{R} = \frac{\lambda}{4\pi\epsilon_0R}$$
$$dE_y = dEsin\theta = k\lambda \frac{x(dx)}{(R^2 + x^2)^{3/2}}$$
$$E_y = k\lambda \int_{0}^{\infty}\frac{x}{(R^2 + x^2)^{3/2}}dx = \frac{k\lambda}{R} = \frac{\lambda}{4\pi\epsilon_0 R}$$

$$\vec E = \frac{\lambda}{4\pi\epsilon_0 R}\hat i + \frac{\lambda}{4\pi\epsilon_0 R}\hat j$$
$$|\vec E| = \frac{\sqrt{2}\lambda}{4\pi\epsilon_0R}$$
The force on the ring due to the field produced by thr rod is hence:
$$F = qE = \frac{\sqrt{2}q\lambda}{4\pi\epsilon_0R}$$
so if my math is correct I just disproved Newtons 3rd law :wink: I did the integrals twice by hand and once using wolfram alpha but still haven't been able to catch the error. Can someone help me out. I really hope there is a mistake in the math and not something conceptually wrong.
 
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  • #2
Your second drawing is confusing. You have ##dE_x## pointing radially out and then element ##dx## along the wire perpendicular to ##dE_x##. Be consistent. That may be your problem because the component of the electric field in the plane of the ring must integrate to zero by symmetry.
 
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  • #3
kuruman said:
Your second drawing is confusing. You have ##dE_x## pointing radially out and then element ##dx## along the wire perpendicular to ##dE_x##. Be consistent. That may be your problem because the component of the electric field in the plane of the ring must integrate to zero by symmetry.
The net electric field in the plane of the ring will have both a x and y component because the rod is not infinitely long in both direction, the bottom of the rod is at the center of the ring and the other end tends to infinity(in the +y direction).
Edit: I think I understand where I went wrong thanks a lot! the net force acting on the ring will only be in the y-direction and the x-componet gets canceled out due to the circular symmetry. all that time spent over nothingo:)
 
  • #4
Hamiltonian299792458 said:
$$E_y = k\lambda \int_{0}^{\infty}\frac{x}{(R^2 + x^2)^{3/2}}dx = \frac{k\lambda}{R} = \frac{\lambda}{4\pi\epsilon_0 R}$$
Everything looks good up until here, although this should be ##E_x## according to your diagram. The only force is in the ##x## direction, as the force in the radial direction cancels out around the ring.
 
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FAQ: Force between charged Ring and Rod, disproved Newtons 3rd law?

What is the force between a charged ring and rod?

The force between a charged ring and rod is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

How does this force disprove Newton's 3rd law?

Newtons's 3rd law states that for every action, there is an equal and opposite reaction. However, in the case of a charged ring and rod, the force acting on the ring is not equal and opposite to the force acting on the rod. This disproves Newton's 3rd law.

Can you explain why the force is not equal and opposite in this scenario?

The force between two charged objects is a result of the interaction between their electric fields. In the case of a ring and rod, the electric fields are not symmetrical, resulting in a net force that is not equal and opposite.

Does this mean that Newton's 3rd law is not valid in all situations?

No, Newton's 3rd law is still valid in most situations. It is a fundamental law of physics and has been proven to hold true in countless experiments. However, in the case of charged objects, the force is not always equal and opposite, as seen in the example of a ring and rod.

Are there any other examples where Newton's 3rd law does not apply?

Yes, there are other scenarios where Newton's 3rd law does not apply, such as in the case of non-contact forces, like magnetic or gravitational forces. In these cases, the forces are not always equal and opposite, but they still follow their respective laws, such as the inverse square law for gravitational forces.

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