A potential field in free space is given as \(V=100 \ln \tan (\theta / 2)+50 \mathrm{~V}\). \((a)\) Find the maximum value of \(\left|\mathbf{E}_{\theta}\right|\) on the surface \(\theta=40^{\circ}\) for \(0.1

Recall that the electric field is given by the negative gradient of the potential field. In spherical coordinates, this means that: $$ \mathbf{E} = -\nabla V = -\left(\frac{\partial V}{\partial r}\hat{\mathbf{r}} + \frac{1}{r}\frac{\partial V}{\partial \theta}\hat{\boldsymbol\theta} + \frac{1}{r\sin\theta}\frac{\partial V}{\partial \phi}\hat{\mathbf{\phi}}\right) $$ Since the potential field depends only on \(\theta\), we only need to find the \(\theta\) component of the electric field: $$ E_{\theta} = -\frac{1}{r}\frac{\partial V}{\partial \theta} $$ Differentiating the potential field with respect to \(\theta\), we get: $$ \frac{\partial V}{\partial \theta} = 100 \frac{\partial}{\partial \theta} \ln \tan(\frac{\theta}{2}) = 100 \frac{1}{\tan(\frac{\theta}{2})}\frac{\partial}{\partial \theta}\tan(\frac{\theta}{2}) = 100 \frac{1}{\tan(\frac{\theta}{2})} \cdot \frac{1}{2} \cdot \sec^2(\frac{\theta}{2}) $$ Hence, we get: $$ E_{\theta} = -\frac{1}{r}\left(100 \frac{1}{\tan(\frac{\theta}{2})} \cdot \frac{1}{2} \cdot \sec^2(\frac{\theta}{2})\right) $$

We are looking for the maximum value of \(|\mathbf{E}_{\theta}|\) on the surface \(\theta = 40^{\circ}\), \(0.1 < r < 0.8m\), and \(60^{\circ} < \phi < 90^{\circ}\). First, let's plug in the value of \(\theta = 40^{\circ}\) into the expression we derived for \(E_{\theta}\): $$ E_{\theta} = -\frac{1}{r}\left(100 \frac{1}{\tan(20^\circ)} \cdot \frac{1}{2} \cdot \sec^2(20^\circ)\right) $$ Notice that \(E_{\theta}\) only depends on \(r\), and it is inversely proportional to \(r\). Therefore, the maximum value of \(|\mathbf{E}_{\theta}|\) will occur at the smallest value of \(r\) in the given range, which is \(r = 0.1m\). Plugging in this value of \(r\), we can find the maximum value of \(|\mathbf{E}_{\theta}|\): $$ |\mathbf{E}_{\theta}|_{max} = 10\left(100 \frac{1}{\tan(20^\circ)} \cdot \frac{1}{2} \cdot \sec^2(20^\circ)\right) $$ Calculating, we get: $$ |\mathbf{E}_{\theta}|_{max} \approx 301.7 \mathrm{~V/m} $$

To describe the surface corresponding to the potential \(V = 80V\), simply set the potential field expression equal to 80V and solve for \(\theta\): $$ 80 = 100 \ln \tan (\frac{\theta}{2}) + 50 $$ Subtract 50 from both sides and divide by 100: $$ \ln \tan (\frac{\theta}{2}) = \frac{3}{10} $$ Take the exponential of both sides to get: $$ \tan (\frac{\theta}{2}) = e^{\frac{3}{10}} $$ Finally, take the inverse tangent of both sides and multiply by 2 to find \(\theta\): $$ \theta = 2\arctan(e^{\frac{3}{10}}) $$ So, the surface is described by the equation \(\theta = 2\arctan(e^{\frac{3}{10}}) \approx 56.26^{\circ}\) with no restrictions on \(r\) and \(\phi\).