带相反电荷的同心导体壳的电势与电场分布咨询

First, let's define our variables clearly to keep things straight:

• Inner conducting shell: radius $a$, carries charge $+Q$ ($Q > 0$)
• Outer conducting shell: inner radius $b$, outer radius $c$ (so $a < b < c$), carries charge $-Q'$ where $Q' > Q$ (since the inner shell has a smaller absolute charge magnitude)

We'll lean on Gauss's Law for electric fields first, then integrate the field to find potentials—symmetry is our biggest ally here.

Electric Field in Each Region #

Gauss's Law states $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}$. The spherical symmetry means $\vec{E}$ is radial and depends only on $r$.

1. Inner Shell Interior ($r < a$) #

For any Gaussian surface inside the inner conductor:

• The enclosed charge is $0$ (all charge on a conductor resides on its surface in electrostatic equilibrium)
• Electrostatic equilibrium also requires the electric field inside a conductor to be zero (free charges would move until the field cancels out entirely)

So:
$$E(r) = 0 \quad (r < a)$$

2. Between the Shells ($a < r < b$) #

A Gaussian surface of radius $r$ here encloses only the inner shell's charge $+Q$. Plugging into Gauss's Law:
$$E(r) \cdot 4\pi r^2 = \frac{Q}{\epsilon_0}$$
Solving for $E(r)$ gives:
$$E(r) = \frac{Q}{4\pi\epsilon_0 r^2} \quad (a < r < b)$$
Direction: Radially outward (since the inner shell is positively charged)

3. Outside the Outer Shell ($r > c$) #

A Gaussian surface here encloses the total charge of both shells: $Q - Q'$. Since $Q' > Q$, the total charge is negative, so the field points radially inward. Applying Gauss's Law:
$$E(r) \cdot 4\pi r^2 = \frac{Q - Q'}{\epsilon_0}$$
The magnitude (with sign to indicate direction) is:
$$E(r) = \frac{Q - Q'}{4\pi\epsilon_0 r^2} \quad (r > c)$$

Potential in Each Region #

Potential is defined as $V(r) = \int_{r}^{\infty} \vec{E} \cdot d\vec{l}$. We'll integrate the electric field piecewise across each region.

1. Inner Shell Interior ($r \leq a$) #

Since the electric field inside the inner shell is zero, the potential at any point inside equals the potential at the inner shell's surface ($r=a$). We integrate from $a$ to infinity, splitting the path into three segments:
$$V(r) = \int_{a}^{b} \frac{Q}{4\pi\epsilon_0 r'^2} dr' + \int_{b}^{c} 0 dr' + \int_{c}^{\infty} \frac{Q - Q'}{4\pi\epsilon_0 r'^2} dr'$$
Calculating each integral and combining terms:
$$V(r) = \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{a} - \frac{1}{b} + \frac{1}{c} \right) - \frac{Q'}{4\pi\epsilon_0 c} \quad (r \leq a)$$

2. Between the Shells ($a < r < b$) #

Integrate from $r$ to infinity, again splitting into regions:
$$V(r) = \int_{r}^{b} \frac{Q}{4\pi\epsilon_0 r'^2} dr' + \int_{b}^{c} 0 dr' + \int_{c}^{\infty} \frac{Q - Q'}{4\pi\epsilon_0 r'^2} dr'$$
Simplifying the result:
$$V(r) = \frac{Q}{4\pi\epsilon_0 r} - \frac{Q}{4\pi\epsilon_0 b} + \frac{Q - Q'}{4\pi\epsilon_0 c} \quad (a < r < b)$$

3. Outside the Outer Shell ($r > c$) #

Here, we only need to integrate the outer region's electric field:
$$V(r) = \int_{r}^{\infty} \frac{Q - Q'}{4\pi\epsilon_0 r'^2} dr'$$
Which evaluates to:
$$V(r) = \frac{Q - Q'}{4\pi\epsilon_0 r} \quad (r > c)$$

内容的提问来源于stack exchange，提问作者Dannnnnnn