Given the potential field \(V=100 x z /\left(x^{2}+4\right) \mathrm{V}\) in free space: \((a)\) Find \(\mathrm{D}\) at the surface \(z=0 .(b)\) Show that the \(z=0\) surface is an equipotential surface. ( \(c\) ) Assume that the \(z=0\) surface is a conductor and find the total charge on that portion of the conductor defined by \(0

Since V is given by \(V = 100\frac{xz}{x^2+4}V\), at z=0, the electric potential V reduces to 0. Using the equation, V, we can write \(V = \frac{ \vec{D} \cdot \vec{dA}}{4\pi\epsilon}\). Now rearranging the equation for D: \(\vec{D} = 4\pi\epsilon \frac{dV}{dz} \hat{z}\). Now, let's differentiate \(V\) with respect to z: \(\frac{dV}{dz} = \frac{100x}{x^2+4}\). Now, substitute the expression for \(\frac{dV}{dz}\) in the D equation: \(\vec{D} = 4\pi\epsilon \frac{100x}{x^2+4} \hat{z}\).

An equipotential surface is a surface that has the same potential (voltage) at every point. To show that the z=0 surface is an equipotential surface, we need to show that the electric potential V is constant along this plane. Considering the given equation for electric potential V: \(V = 100\frac{xz}{x^2+4}V\). On the z=0 plane, V becomes: \(V = 100\frac{x*0}{x^2+4}V = 0\). Since the electric potential V is constant (0) along the z=0 plane, it can be concluded that the z=0 surface is an equipotential surface.

To find the total charge on the conductor, we need to integrate the charge density over the area defined by the given \(x\) and \(y\) limits. First, let's find the charge density \(\rho_s\): \(\rho_s = \hat{z} \cdot \vec{D} = 4\pi\epsilon \frac{100x}{x^2+4}\). Now, integrate over the given area: \(Q = \int_{0}^{2} \int_{-3}^{0} \rho_s dxdy\). \(Q = 4\pi\epsilon \int_{0}^{2} \int_{-3}^{0} \frac{100x}{x^2+4} dydx\). Since the integral is not a function of y, the integral splits into two parts: \(Q = 4\pi\epsilon (-3) \int_{0}^{2} \frac{100x}{x^2+4} dx\). Now integrate with respect to x: \(Q = -12\pi\epsilon \left[\frac{50}{2} \tan^{-1}\frac{x}{2} \right]_{0}^{2}\). \(Q = -12\pi\epsilon \left[\frac{50}{2}(\tan^{-1}(1)-\tan^{-1}(0))\right]\). \(Q = -300\pi\epsilon \left[\frac{\pi}{4}\right]\). So the total charge on the conductor is: \(Q = -75\pi^2\epsilon\).