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

Step-by-step 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) \\&=(2 \pi \sigma)/(4 \pi \varepsilon_(0))[\sqrt{z^(2)+R^(2)}-(21)] \\&=(\sigma)/(2)(\sqrt{2^(2)+R^(2)}-2)\end{aligned}`

Note: Refer the image attached