Answer:
E=![\frac{KQ}{2\sqrt 2a^2}](https://tex.z-dn.net/?f=%5Cfrac%7BKQ%7D%7B2%5Csqrt%202a%5E2%7D)
Explanation:
We are given that
Charge on ring= Q
Radius of ring=a
We have to find the magnitude of electric filed on the axis at distance a from the ring's center.
We know that the electric field at distance x from the center of ring of radius R is given by
![E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BkQx%7D%7B%28R%5E2%2Bx%5E2%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D)
Substitute x=a and R=a
Then, we get
![E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BKQa%7D%7B%28a%5E2%2Ba%5E2%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D)
![E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BKQa%7D%7B%282a%5E2%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D)
![E=\frac{KQa}{2\sqrt 2a^3}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BKQa%7D%7B2%5Csqrt%202a%5E3%7D)
![E=\frac{KQ}{2\sqrt 2a^2}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BKQ%7D%7B2%5Csqrt%202a%5E2%7D)
Where K=![9\times 10^9 Nm^2/C^2](https://tex.z-dn.net/?f=9%5Ctimes%2010%5E9%20Nm%5E2%2FC%5E2)
Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=![\frac{KQ}{2\sqrt 2a^2}](https://tex.z-dn.net/?f=%5Cfrac%7BKQ%7D%7B2%5Csqrt%202a%5E2%7D)
The U.S. Environmental Protection Agency (EPA) standard for nitrate in drinking water is 10 milligrams of nitrate (measured as nitrogen) per liter of drinking water (mg/L). * Drinking water with levels of nitrate at or below 10 mg/L is considered safe for everyone.
Could be easy for some people and hard for some people.