Why Is the Electric Potential the Same for Inner and Outer Semi-Circles?

[cherry]

Homework Statement

Two semicircular rods and two short, straight rods are joined in the configuration shown. The rods carry a charge of λ = 55C/m. The radii are R1 = 4.3 m and R2 = 9.5 m.

Calculate the potential at the centre of this configuration (point P). Please note that there are also charges on the straight segments.

Relevant Equations

V = kλ ln[r + l / r]
V = kq / r

I solved using the formulae listed in the relevant equations and got the right answer.
However, I noticed something strange to me.
The electric potential due to the inner semi-circle was equal to that due to the outer semi-circle.
But based on the formula for calculating V, we notice that there is "r" involved.
So shouldn't the electric potential be different?

 

[cherry]

My professor gave a hint saying to "utilize spherical symmetry" but I still don't understand why they would be equal.

 

[haruspex]

the configuration shown.

?

 

[cherry]

?

Hi! Sorry, I forgot to upload the picture. For some reason, I am not getting the settings to upload a photo. So here is a Google Drive image of the diagram.

 

[Delta2]

Homework Statement: Two semicircular rods and two short, straight rods are joined in the configuration shown. The rods carry a charge of λ = 55C/m. The radii are R1 = 4.3 m and R2 = 9.5 m.

Calculate the potential at the centre of this configuration (point P). Please note that there are also charges on the straight segments.
Relevant Equations: V = kλ ln[r + l / r]
V = kq / r

The electric potential due to the inner semi-circle was equal to that due to the outer semi-circle.

The qualitative explanation is that the inner semi-circle has smaller radius but has smaller semi-circumference too, so smaller total charge. It happens that the total charge of each of the two semi-circumference , is equal to $q_1=\pi \lambda r_1$, $q_2=\pi \lambda r_2$ while the potentials (at center) are $V_1=K\frac{q_1}{r_1}$ and $V_2=K\frac{q_2}{r_2}$ , so you can see that each of the radius $r_1,r_2$ gets simplified when calculating the potentials and leave just $\pi K\lambda$ in both cases

 

[cherry]

So whenever there are concentric circles, the electric potential is the same?

The qualitative explanation is that the inner semi-circle has smaller radius but has smaller semi-circumference two, so smaller total charge. It happens that the total charge of each of the two semi-circumference , is equal to $q_1=\pi \lambda r_1$, $q_2=\pi \lambda r_2$ while the potentials (at center) are $V_1=K\frac{q_1}{r_1}$ and $V_2=K\frac{q_2}{r_2}$ , so you can see that each of the radius $r_1,r_2$ gets simplified when calculating the potentials and leave just $\pi K\lambda$ in both cases

Oh, so whenever there are concentric circles with different radii, the electric potential is the same across all circles?

 

[Delta2]

So whenever there are concentric circles, the electric potential is the same?

Oh, so whenever there are concentric circles with different radii, the electric potential is the same across all circles?

Yes. given that all concentric circles carry the same linear charge density $\lambda$.

And also I think this holds only for the potential at the center of the configuration.

 

[cherry]

Yes. given that all concentric circles carry the same linear charge density $\lambda$.

And also I think this holds only for the potential at the center of the configuration.

Got it, thank you!