Question

Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by integrating the electric field e along the axis of the disk

Answer 1

Explanation:

Area of ring $\ 2{\pi} a d a$

Charge of on ring $d q=-(\ 2{\pi} a d a)$

Charge on disk

$Q=-\left(\pi R^{2}\right)$

$\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}$

Note: Refer the image attached