Question

What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.12 m whose potential is 230 V (with V = 0 at infinity)?

Answer 1

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:

$V_{potential}=k_{constat}\frac{q_{charge}}{r_{radius}} \\\\q_{charge}=\frac{V_{potential*r_{radius}}}{k_{constant}}$

Substitute the given values

$q_{charge}=\frac{(0.12m)(230V)}{9*10^{9} }\\ q_{charge}=3.07*10^{-9}C\\or\\q_{charge}=3.07nC$

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

$=\frac{3.07*10^{-9}C}{4\pi (0.12m)^2}\\ =1.69*10^{-8}C/m^2\\or\\=17nC/m^2$

σ=17 nC/m²