Question

A long wire with a uniform linear charge density $\lambda$ is placed along the common axis of two long coaxial thin cylindrical shells of radii $R_1$ and $R_2$ (see figure). The outer shell carries a uniform surface charge density $-\lambda/(2\pi R_2)$. Take the zero of the potential at infinity along the radial direction. The potential at a distance $r$ from the axis where $R_1 < r < R_2$ is

• A. $\frac{\lambda}{4\pi\epsilon_0}\ln\left(\frac{R_2}{r}\right)$
• B. $\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{R_2}{r}\right)$
• C. $\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{r}{R_2}\right)$
• D. $\frac{\lambda}{4\pi\epsilon_0}\ln\left(\frac{r}{R_2}\right)$

Answer 1

Answer: (B)

$\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{R_2}{r}\right)$

Answer 2

The problem asks for the electric potential at a distance $r$ from the axis, where $R_1 < r < R_2$. The system consists of a long wire with linear charge density $\lambda$ and an outer cylindrical shell of radius $R_2$ with surface charge density $-\lambda/(2\pi R_2)$. The inner shell at radius $R_1$ is mentioned but no charge is specified on it, so we assume it is uncharged and does not affect the field outside it. The potential is set to zero at infinity along the radial direction.

1. Determine the Electric Field in different regions using Gauss's Law:

• For $R_1 < r < R_2$:

  Consider a cylindrical Gaussian surface of radius $r$ and length $L$, coaxial with the wire. The only charge enclosed by this surface is from the central wire, which is $Q_{enclosed} = \lambda L$.

  According to Gauss's Law, $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0}$.

  Due to cylindrical symmetry, the electric field $\vec{E}$ is radial and its magnitude is constant on the Gaussian surface.

  $E (2\pi r L) = \frac{\lambda L}{\epsilon_0}$

  Therefore, the electric field in this region is:

  $E(r) = \frac{\lambda}{2\pi\epsilon_0 r}$

• For $r > R_2$:

  Consider a cylindrical Gaussian surface of radius $r$ and length $L$, coaxial with the wire. This surface encloses the central wire and the outer cylindrical shell.

  The charge on the central wire is $Q_{wire} = \lambda L$.

  The total charge on a length $L$ of the outer shell is $Q_{outer} = (\text{surface charge density}) \times (\text{surface area})$

  $Q_{outer} = \left(-\frac{\lambda}{2\pi R_2}\right) (2\pi R_2 L) = -\lambda L$.

  The total charge enclosed by the Gaussian surface is $Q_{total} = Q_{wire} + Q_{outer} = \lambda L - \lambda L = 0$.

  According to Gauss's Law, $E (2\pi r L) = \frac{0}{\epsilon_0}$, which implies:

  $E(r) = 0$ for $r > R_2$.

2. Determine the Reference Potential:

The problem states that the potential at infinity along the radial direction is zero, i.e., $V(\infty) = 0$.

Since the electric field $E(r) = 0$ for all $r > R_2$, the potential $V(r)$ must be constant for $r > R_2$.

Given $V(\infty) = 0$, it follows that $V(r) = 0$ for all $r \ge R_2$.

Therefore, the potential at the outer shell's radius is $V(R_2) = 0$.

3. Calculate the Potential at a distance $r$ where $R_1 < r < R_2$:

The potential difference between two points $r_a$ and $r_b$ is given by $V(r_b) - V(r_a) = -\int_{r_a}^{r_b} \vec{E} \cdot d\vec{l}$.

We want to find $V(r)$ where $R_1 < r < R_2$. We can integrate from $R_2$ (where $V(R_2)=0$) to $r$.

$V(r) - V(R_2) = -\int_{R_2}^{r} E(r') dr'$

Substitute $V(R_2) = 0$ and $E(r') = \frac{\lambda}{2\pi\epsilon_0 r'}$:

$V(r) - 0 = -\int_{R_2}^{r} \frac{\lambda}{2\pi\epsilon_0 r'} dr'$

$V(r) = -\frac{\lambda}{2\pi\epsilon_0} [\ln r']_{R_2}^{r}$

$V(r) = -\frac{\lambda}{2\pi\epsilon_0} (\ln r - \ln R_2)$

Using the logarithm property $\ln a - \ln b = \ln(a/b)$:

$V(r) = -\frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{r}{R_2}\right)$

Using another logarithm property $-\ln(x) = \ln(1/x)$:

$V(r) = \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{R_2}{r}\right)$

Comparing this result with the given options, it matches option (B).