Let \(V=10(\rho+1) z^{2} \cos \phi \mathrm{V}\) in free space. \((a)\) Let the equipotential surface \(V=20 \mathrm{~V}\) define a conductor surface. Find the equation of the conductor surface. \((b)\) Find \(\rho\) and \(\mathbf{E}\) at that point on the conductor surface where \(\phi=\) \(0.2 \pi\) and \(z=1.5 .(c)\) Find \(\left|\rho_{S}\right|\) at that point.

Given that the equipotential surface is defined for \(V = 20V\), we plug this into the given expression for \(V\) to find the equation of the conductor surface: \(20 = 10(\rho + 1)z^2 \cos\phi\). Divide both sides by 10, we obtain: \(2 = (\rho + 1)z^2 \cos\phi\).

The electric field (\(\mathbf{E}\)) is related to the voltage via \(\mathbf{E} = -\nabla V\). So, we need to first calculate the gradient of \(V\) to find the electric field. We are given \(V = 10(\rho+1) z^{2} \cos \phi\), so: \(\frac{\partial V}{\partial \rho} = 10z^2 \cos\phi\) \(\frac{\partial V}{\partial \phi} = -10(\rho+1)z^2 \sin\phi\) \(\frac{\partial V}{\partial z} = 20(\rho+1)z \cos\phi\) Thus, the electric field is: \(\mathbf{E} = -\left(\frac{\partial V}{\partial \rho}\boldsymbol{\hat{\rho}} + \frac{\partial V}{\partial \phi}\boldsymbol{\hat{\phi}} + \frac{\partial V}{\partial z}\boldsymbol{\hat{z}}\right)\)

Since the problem is in free space, volume charge density \(\rho = 0~C/m^3\).

To find the surface charge density \(|\rho_S|\), we need to first evaluate the electric field at the given point \((\phi = 0.2\pi, z = 1.5)\) on the conductor surface. Plug these values into the equation for the conductor surface from Step 1: \(2 = (\rho + 1)(1.5)^2 \cos(0.2\pi)\) Now, we can find the corresponding \(\rho\) value: \(\rho = \frac{2}{(1.5)^2\cos(0.2\pi)} - 1\) Now, plug \(\rho\), \(z\), and \(\phi\) into the electric field expression found in Step 2: \(\mathbf{E} = -\left(10(1.5)^2\cos(0.2\pi)\boldsymbol{\hat{\rho}} - 10(\rho+1)(1.5)^2\sin(0.2\pi)\boldsymbol{\hat{\phi}} + 20(\rho+1)(1.5)\cos(0.2\pi)\boldsymbol{\hat{z}}\right)\) Since the point is on the conductor surface, the tangential component of the electric field is zero. However, the normal component of the electric field is related to the surface charge density \(\rho_S\) via Gauss' law: \(\mathbf{E}_n = \frac{\rho_S}{\varepsilon_0}\). Therefore, we can find \(|\rho_S|\) by multiplying the normal component of the electric field by the vacuum permittivity, \(\varepsilon_0\) (approximately \(8.85\times10^{-12}C^2/Nm^2\)): \(|\rho_S| = \varepsilon_0|\mathbf{E}_n|\)