Question

The potential of a 5.0 cm radius conducting sphere is -100 V. What is the charge density on its surface?Group of answer choices-1.8x10-8 C/m23.5x10-8 C/m22.2x10-8 C/m2-2.2x10-8 C/m2-3.5x10-8 C/m2

Answer 1

First lets calculate the surface area

`S=4\cdot\pi\cdot r^2=4\pi\cdot0.05^2m=0.0314m^2`

Now to know the surface charge density we need the next formula:

`CD=(q)/(A)`

But we are missing the amount of charge, we only have the potential

So in this case, we going to apply a different formula

`V=(q)/(4\cdot\pi\cdot\xi\cdot r)`

q=5.56*10^-10

`CD=-5.56\cdot(10^(-10))/(0.0314)=-1.77\cdot(10^(-8)C)/(m^2)`

The anwer might be -1.8x10-8