Answer:
(a) q=3.07 nC
(b) σ=17 nC/m²
Explanation:
Given data
Radius r=0.12m
Potential V=230 V
To find
(a) Charge q
(b) Charge density σ
Solution
For Part (a)
As we know that potential is:
[tex]V_{potential}=k_{constat}\frac{q_{charge}}{r_{radius}} \\\\q_{charge}=\frac{V_{potential*r_{radius}}}{k_{constant}}[/tex]
Substitute the given values
[tex]q_{charge}=\frac{(0.12m)(230V)}{9*10^{9} }\\ q_{charge}=3.07*10^{-9}C\\or\\q_{charge}=3.07nC[/tex]
For Part (b)
The charge density is given by:
σ=q/(4πr²)
Substitute the given values and value of q to find charge density
So
[tex]=\frac{3.07*10^{-9}C}{4\pi (0.12m)^2}\\ =1.69*10^{-8}C/m^2\\or\\=17nC/m^2[/tex]
σ=17 nC/m²
