Two concentric conducting spherical shells of radii $a_1$ and $a_2$ ($a_2>a_1$), are charged to potentials $φ_1$ and $φ_2$, respectively. Determine the electric potential and field in the region between the shells. Also determine the charge on the inner shell..?

Why we need Laplace's equation to solve this problem?

Why I can't just solve it by Gauss law, as the charge on first shell will be $4π ε oa_1φ_1$ (as potential $φ1=q/4π ε oa_1$ and also since the shells are conducting, charges will distribute on the surface of the shells uniformly, so there will be spherical symmetry)?


Consider that the electrostatic potential $\varphi$ isn't directly observable. The potential $\varphi+C$ where $C$ is a constant gives the same electric field, and so the same physics. Because changing $\varphi$ by a constant should give the same physics, you cannot conclude that the charge on the shell is $4\pi\epsilon_0 a_1 \varphi_1$. The charge is a physical, observable quantity, but the value of the potential at a point is not.

To fix this constant you need some boundary condition, for example that $\varphi$ is $0$ at infinity. This boundary condition is the one used when we say that the potential from a charge is proportional to $q/r$. But actually, any $\varphi$ of the form $$\varphi = \frac{q}{4\pi \epsilon_0 r} + C \tag{1}$$ will do. This form of the potential can be found by finding the electric field with Gauss's law and integrating, remembering to add the integration constant. With the potential in the from (1) your boundary conditions are \begin{align} \varphi_1 = \frac{q}{4\pi\epsilon_0 a_1} + C \\ \varphi_2 = \frac{q}{4\pi\epsilon_0 a_2} + C \\ \end{align} Here the unknowns are the charge $q$ and the integration constant $C$, and the two equations for two unknowns can be solved.

You can derive an equivalent system of equations by applying Gauss's law, more similar to your original suggestion. By the usual argument of spherical symmetry, the electric field must be $\mathbf E = \frac{q}{4\pi \epsilon_0 r^2}\hat r$. Integrating along any path from $r = a_1$ to $r = a_2$, we have $$\int_{a_1}^{a_2} \mathbf E \cdot d\mathbf r = \varphi_1 - \varphi_2 = \frac{q}{4\pi\epsilon_0}(-\frac{1}{a_2} + \frac{1}{a_1})$$ which is an equation for $q$. Using that the general potential is of the form (1), we obtain also the constant $C$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.