Answer:
Therefore,
a) ![E=20.82\ kN/C](https://tex.z-dn.net/?f=E%3D20.82%5C%20kN%2FC)
b) ![V= 2290\ Volt](https://tex.z-dn.net/?f=V%3D%202290%5C%20Volt)
c) distance from the sphere's surface = 1.8 cm
Explanation:
Given:
Radius, r = 11 cm = 0.11 m
Charge, ![Q = 2.8\times 10^{-8}\ C](https://tex.z-dn.net/?f=Q%20%3D%202.8%5Ctimes%2010%5E%7B-8%7D%5C%20C)
To Find:
a) electric field at the sphere's surface = ?
b) If V = 0 at infinity, what is the electric potential at the sphere's surface = ?
c) At what distance from the sphere's surface has the electric potential decreased by 320 V = ?
Solution:
Electric field at the sphere's surface is given as,
![E=\dfrac{k\times Q}{r^{2}}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bk%5Ctimes%20Q%7D%7Br%5E%7B2%7D%7D)
Where,
E = Electric Field,
![k = Coulomb's\ constant = 9\times 10^{9}](https://tex.z-dn.net/?f=k%20%3D%20Coulomb%27s%5C%20constant%20%3D%209%5Ctimes%2010%5E%7B9%7D)
Q = Charge
r = Radius
Substituting the values we get
![E=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{(0.11)^{2}}=20.82\ kN/C](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B9%5Ctimes%2010%5E%7B9%7D%5Ctimes%202.8%5Ctimes%2010%5E%7B-8%7D%7D%7B%280.11%29%5E%7B2%7D%7D%3D20.82%5C%20kN%2FC)
Now, Electric Potential at point surface is given as,
![V=\dfrac{k\times Q}{r}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7Bk%5Ctimes%20Q%7D%7Br%7D)
Substituting the values we get
![V=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{0.11}=2.29\ kV](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B9%5Ctimes%2010%5E%7B9%7D%5Ctimes%202.8%5Ctimes%2010%5E%7B-8%7D%7D%7B0.11%7D%3D2.29%5C%20kV)
For distance from the sphere's surface has the electric potential decreased by 320 V,
So V becomes,
V = 2290 - 320 = 1970 Volt, then r =?
![V=\dfrac{k\times Q}{r}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7Bk%5Ctimes%20Q%7D%7Br%7D)
Substituting the values we get
![r=\dfrac{k\times Q}{V}\\\\r=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{1970}=0.128\ m](https://tex.z-dn.net/?f=r%3D%5Cdfrac%7Bk%5Ctimes%20Q%7D%7BV%7D%5C%5C%5C%5Cr%3D%5Cdfrac%7B9%5Ctimes%2010%5E%7B9%7D%5Ctimes%202.8%5Ctimes%2010%5E%7B-8%7D%7D%7B1970%7D%3D0.128%5C%20m)
Therefore the distance from the sphere's surface,
![d = 12.8 - 11 = 1.8\ cm](https://tex.z-dn.net/?f=d%20%3D%2012.8%20-%2011%20%3D%201.8%5C%20cm)
Therefore,
a) ![E=20.82\ kN/C](https://tex.z-dn.net/?f=E%3D20.82%5C%20kN%2FC)
b) ![V= 2290\ Volt](https://tex.z-dn.net/?f=V%3D%202290%5C%20Volt)
c) distance from the sphere's surface = 1.8 cm