Electric potential of charged concentric spheres

In summary, the conversation discusses the calculation of the charge q_1 necessary for the potential on the inner sphere to be zero. Gauss' law is used to determine the electric potential on the surface and inside each sphere, and it is found that q_1 must be equal to -R_1/R_2 times q_2. A proposed solution for q_1=0 is deemed incorrect.
  • #1
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Homework Statement



Two conducting concentric spheres of negligible thickness. The radii of the spheres are [itex]R_1[/itex] and [itex]R_2[/itex], respectively, with [itex]R_2>R_1[/itex]. A charge [itex]q_2[/itex] is placed on the external sphere.
A charge [itex]q_1[/itex] is placed on the internal sphere.

Assume that the electric potential is zero infinitely far from the center of the spheres.

Find [itex]q_1[/itex] such that the potential on the inner sphere is zero.

Homework Equations



  • Gauss law: [itex]\Phi(\vec{E}) = \frac{Q}{\varepsilon_0}[/itex]
  • Potential of a point charge (such that [itex]V=0[/itex] for [itex]d=\infty[/itex]): [itex]\displaystyle V= k \frac{Q}{d} [/itex].

The Attempt at a Solution



Using Gauss' law I can say that, if only the outer sphere was present,
[tex] V_2 = k \frac{q_2}{R_2} [/tex]
on its surface and inside it. Hence, at radius [itex]R_1[/itex],
[tex] V_1 = k \frac{q_2}{R_2} [/tex]
If only the inner sphere was present
[tex] V_1 = k \frac{q_1}{R_1} [/tex]
on its surface
When both spheres are present, on the surface of the inner sphere
[tex] V_1 = k \frac{q_2}{R_2}+k \frac{q_1}{R_1} [/tex]
If we want [itex]V_1=0[/itex], it should be
[tex] \frac{q_2}{R_2}=- \frac{q_1}{R_1} [/tex]
and hence
[tex] q_1=- \frac{R_1}{R_2} q_2 [/tex]

It seems to me that this makes sense. The potential would be
[tex] V_2 = \frac{k} q_2 \left\{
\begin{array}{cc}
\frac{1}{r}\left(1-\frac{R_1}{R_2}\right) & r\geq R_2 \\
\frac{1}{R_2}-\frac{1}{r}\frac{R_1}{R_2} & R_1 \leq r \leq R_2
\end{array}
\right.
[/tex]

Assuming that [itex]q_2>0[/itex], the potential grows when the outer sphere is approached from outside, reaches its maximum on the surface of the spheres and drops to 0 when it reaches the inner radius...
Inside that, it stays zero.I'm asking here just to be on the safe side.
I came across a (kind of "official") solution for this exercise that in my opinion does not make any sense. It concludes [itex]q_1=0[/itex].

[edited]
 
Last edited:
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  • #2
Your solution looks good to me. The "kind of official" solution does not: If the charge on the inner sphere were zero it would have no influence on the potential at its surface, and so the potential there would have to be the same as that of the surface of the outer sphere which is not zero as required by the problem.
 
  • #3
Hi gneill,

thanks for your insight. I was pretty sure that the proposed solution could not be correct. Other than the final result, the entire derivation seems really shaky, but there's no point in repeating a wrong reasoning.
I needed some support because I'm a bit rusty on this stuff, and I was afraid I was missing a subtle point.
Thanks again
 
  • #4
There are online video tutorials that cover this topic which may be of interest. Google: Potential in a System of Concentric Shells
 

1. What is the formula for calculating the electric potential of charged concentric spheres?

The formula for calculating the electric potential of charged concentric spheres is V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge of the inner sphere, and r is the distance between the two spheres.

2. How does the electric potential change as the distance between the spheres increases?

As the distance between the spheres increases, the electric potential decreases. This is because the electric potential is inversely proportional to the distance between the spheres (V ∝ 1/r). Therefore, as r increases, V decreases.

3. Can the electric potential of charged concentric spheres ever be negative?

Yes, the electric potential of charged concentric spheres can be negative. This occurs when the inner sphere has a negative charge and the outer sphere has a positive charge. In this case, the electric potential is negative because the electric field points from the outer sphere to the inner sphere.

4. How does the electric potential of charged concentric spheres compare to that of two point charges?

The electric potential of charged concentric spheres is similar to that of two point charges, except that the electric potential between the spheres is constant. This is because the electric potential of charged concentric spheres depends only on the distance between the spheres, while the electric potential between two point charges also depends on the magnitude of the charges.

5. How can the electric potential of charged concentric spheres be used in practical applications?

The electric potential of charged concentric spheres can be used in various practical applications, such as in voltage regulators and electrostatic precipitators. It is also used in experiments to demonstrate the relationship between electric potential and distance, as well as to calculate the electric field between the spheres.

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