(continued): The magnitude of the electric field dE created by a point charge dq at a distance d = sqrt(r2 + z2) is dE = (1)/(4pi0) (dq)/(d2).

Okay, Aïcha, let's continue with .

Step 1 (continued): The magnitude of the electric field $dE$ created by a point charge $dq$ at a distance $d = \sqrt{r^2 + z^2}$ is $dE = \frac{1}{4\pi\epsilon_0} \frac{dq}{d^2}$. Due to symmetry, only the component along the z-axis, $dE_z$, contributes to the total field. $dE_z = dE \cos\alpha$ where $\cos\alpha = \frac{z}{d} = \frac{z}{\sqrt{r^2 + z^2}}$. So, for the infinitesimal ring: $dE_z = \frac{1}{4\pi\epsilon_0} \frac{dq}{r^2 + z^2} \frac{z}{\sqrt{r^2 + z^2}} = \frac{z}{4\pi\epsilon_0} \frac{dq}{(r^2 + z^2)^{3/2}}$ Substitute $dq = \sigma (2\pi r dr)$: $dE_z = \frac{z}{4\pi\epsilon_0} \frac{\sigma (2\pi r dr)}{(r^2 + z^2)^{3/2}} = \frac{\sigma z r dr}{2\epsilon_0 (r^2 + z^2)^{3/2}}$

Step 2: Integrate $dE_z$ over the entire disk from $r=0$ to $r=R$ to find the total electric field $E_z$. $E_z = \int_{0}^{R} \frac{\sigma z r dr}{2\epsilon_0 (r^2 + z^2)^{3/2}}$ Let $u = r^2 + z^2$. Then $du = 2r dr$, so $r dr = \frac{1}{2} du$. When $r=0$, $u = z^2$. When $r=R$, $u = R^2 + z^2$. $E_z = \int_{z^2}^{R^2 + z^2} \frac{\sigma z}{2\epsilon_0} \frac{1}{u^{3/2}} \frac{1}{2} du = \frac{\sigma z}{4\epsilon_0} \int_{z^2}^{R^2 + z^2} u^{-3/2} du$

Step 3: Evaluate the integral. $E_z = \frac{\sigma z}{4\epsilon_0} \left[ \frac{u^{-1/2}}{-1/2} \right]_{z^2}^{R^2 + z^2} = \frac{\sigma z}{4\epsilon_0} \left[ \frac{-2}{\sqrt{u}} \right]_{z^2}^{R^2 + z^2}$ $E_z = \frac{\sigma z}{4\epsilon_0} \left( \frac{-2}{\sqrt{R^2 + z^2}} - \frac{-2}{\sqrt{z^2}} \right)$ $E_z = \frac{\sigma z}{4\epsilon_0} \left( \frac{2}{|z|} - \frac{2}{\sqrt{R^2 + z^2}} \right)$

Step 4: Simplify the expression. Assuming $z > 0$ (point M is on the positive z-axis), then $|z|=z$. $E_z = \frac{\sigma z}{4\epsilon_0} \left( \frac{2}{z} - \frac{2}{\sqrt{R^2 + z^2}} \right)$ $E_z = \frac{\sigma z}{2\epsilon_0} \left( \frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2}} \right)$ $E_z = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{z}{\sqrt{R^2 + z^2}} \right)$ The electric field vector is along the z-axis: \vec{E}(z) = \frac{\sigma{2\epsilon_0} \left( 1 - \frac{z}{\sqrt{R^2 + z^2}} \right) \vec{k}}