Solution :
We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :




If we put an electron on point P, then force on point e is :

![F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7B-eKQx%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B3%2F2%7D%7D%3D%20%5Cfrac%7B-eKQx%7D%7Br%5E3%5B1%2B%5Cfrac%7Bx%5E2%7D%7Br%5E2%7D%5D%5E%7B3%2F2%7D%7D)
If r >> x , then 
Then, 


Compare, a = -ω²x
We get,




The next step is -748 divide by -11 is 68 m (answer) the pic got cropped sorry
Sample Response: How can magnetic and electric fields be demonstrated?
Reflection from such a rough surface is called diffuse reflection and appears matte
Answer:
α = τ/I = 0.77 / (0.70(0.30²)) = 12.22222... = 12 rad/s²
Explanation: