Let's go step by step for each part: --- **a) Find a generic expression for the electric field \( E_r \) inside the Geiger counter as a function of radial distance \( r \).** **Given:** - Central wire: radius \( r_w \) - Outer cylinder: radius \( r_c \) - Charge per unit length on wire: \( \lambda \) - Charge per unit length on cylinder: \( -\lambda \) **Approach:** The system is a coaxial cable (cylinder) with a line charge in the center. We can use **Gauss's Law** to find the electric field at a distance \( r \) from the axis, where \( r_w < r < r_c \): \[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0} \] The Gaussian surface is a cylinder of length \( L \) and radius \( r \), coaxial with the wire. - The charge enclosed: \( q_{enc} = \lambda L \) - The area through which \( \vec{E} \) passes: \( A = 2\pi r L \) - The electric field is radial and uniform over this surface. So, \[ E_r(2\pi r L) = \frac{\lambda L}{\varepsilon_0} \implies E_r = \frac{\lambda}{2\pi \varepsilon_0 r} \] **Answer for (a):** \[ \boxed{ E_r(r) = \frac{\lambda}{2\pi \varepsilon_0 r} \quad \text{for} \quad r_w < r < r_c } \] --- **b) General expression for the potential difference \( \Delta V_{wc} \) between the wire and the cylinder** **Given:** - Inner wire radius: \( r_w \) - Outer cylinder radius: \( r_c \) - Electric field: \( E_r(r) = \frac{\lambda}{2\pi\varepsilon_0 r} \) **Approach:** Potential difference is the (negative) integral of the electric field from the cylinder to the wire: \[ \Delta V_{wc} = V_w - V_c = -\int_{r_c}^{r_w} E_r(r) \, dr \] Plug in \( E_r(r) \): \[ \Delta V_{wc} = -\int_{r_c}^{r_w} \frac{\lambda}{2\pi\varepsilon_0 r} \, dr \] \[ = -\frac{\lambda}{2\pi\varepsilon_0} \int_{r_c}^{r_w} \frac{1}{r} dr \] \[ = -\frac{\lambda}{2\pi\varepsilon_0} [\ln r]_{r_c}^{r_w} \] \[ = -\frac{\lambda}{2\pi\varepsilon_0} (\ln r_w - \ln r_c) \] \[ = -\frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_w}{r_c}\right) \] But since the inner wire is at higher potential, the potential difference is positive (potential drops from wire to cylinder), so: \[ \Delta V_{wc} = \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_c}{r_w}\right) \] --- **Answer for (b):** \[ \boxed{ \Delta V_{wc} = \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_c}{r_w}\right) } \] --- **Summary:** - **(a)** \( E_r(r) = \displaystyle \frac{\lambda}{2\pi\varepsilon_0 r} \) - **(b)** \( \Delta V_{wc} = \displaystyle \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_c}{r_w}\right) \)