Uniform surface charge densities of 6 and \(2 \mathrm{nC} / \mathrm{m}^{2}\) are present at \(\rho=2\) and \(6 \mathrm{~cm}\), respectively, in free space. Assume \(V=0\) at \(\rho=4 \mathrm{~cm}\), and calculate \(V\) at \((a) \rho=5 \mathrm{~cm} ;(b) \rho=7 \mathrm{~cm} .\)

Given that we are working in cylindrical coordinates, we can use Gauss's Law to find the electric field due to the surface charge densities: \(E_\rho = \frac{\sigma}{2\pi\epsilon_0\rho}\) For the surface charge densities \(\sigma_1 = 6\) nC/m² and \(\sigma_2 = 2\) nC/m² at \(\rho_1 = 2\) cm and \(\rho_2 = 6\) cm, respectively, we can find the corresponding electric field magnitudes in each case. \(E_{\rho_1} = \frac{6 \times 10^{-9}}{2\pi\times 8.85 \times 10^{-12} \times 2 \times 10^{-2}}\) \(E_{\rho_2} = \frac{2 \times 10^{-9}}{2\pi\times 8.85 \times 10^{-12} \times 6 \times 10^{-2}}\) After solving, we get the following electric field magnitudes: \(E_{\rho_1} \approx 53.93 \, \mathrm{V/m}\) \(E_{\rho_2} \approx 5.99 \, \mathrm{V/m}\)

We are given that the electric potential at \(\rho=4\) cm is zero, i.e. \(V(4) = 0\). Since the electric field is radial in nature, we will consider the radial coordinate only when integrating the electric field over a distance. From \(2\) cm to \(4\) cm, only the surface charge density at \(\rho_1\) contributes. Therefore, the integration will be: \(0 - V_1 = -\int_{2 \times 10^{-2}}^{4 \times 10^{-2}} E_{\rho_1} \,d\rho\) Solving for \(V_1\), we get: \(V_1 = -53.93 \times (4 \times 10^{-2} - 2 \times 10^{-2}) \approx -107.86 \, \mathrm{V}\)

To find the electric potential at \(\rho=5\) cm, we will integrate the electric field again: \(V(5) = V_1 + \int_{4 \times 10^{-2}}^{5 \times 10^{-2}} (-E_{\rho_1} + E_{\rho_2}) \,d\rho\) After plugging the values of \(E_{\rho_1}\) and \(E_{\rho_2}\), we get: \(V(5) = -107.86 + \int_{0.04}^{0.05} (-53.93 + 5.99) \,d\rho\) Solving for \(V(5)\), we obtain: \(V(5) \approx -128.04 \, \mathrm{V}\)

Similarly, to find the electric potential at \(\rho=7\) cm, we will integrate the electric field: \(V(7) = V(5) + \int_{5 \times 10^{-2}}^{7 \times 10^{-2}} (-E_{\rho_1} - E_{\rho_2}) \,d\rho\) After plugging the values of \(E_{\rho_1}\) and \(E_{\rho_2}\), we get: \(V(7) = -128.04 + \int_{0.05}^{0.07} (-53.93 - 5.99) \,d\rho\) Solving for \(V(7)\), we obtain: \(V(7) \approx -166.00 \, \mathrm{V}\) Thus, the electric potential at \(\rho=5\) cm is approximately \(-128.04\) V, and the electric potential at \(\rho=7\) cm is approximately \(-166.00\) V.