Question

A uniformly charged conducting sphere of 0.93 m diameter has a surface charge density of 9.5 µC/m^2. (a) Find the net charge in coulombs on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Answer 1

For part (a) I got an answer of .000056 C...
part (b) I am unsure.

Answer 2

Again use Gauss's law

Answer 3

"flux x area = q / epsilonsub0" @douglaswinslowcooper
this is wrong :-/

Answer 4

For the first part, I did that. Except I forgot the squared.

\(\Large{q=4\cdot\pi\cdot(.465)^2=2.71716}\)

Then I just use,

\(\Large{\Phi=\frac{q}{\epsilon_{0}}=\frac{2.71716}{8.85\times10^{-12}}=3.07024\times10^{11}}\)

Of course, sig figs are off.

Answer 5

They have asked total electric flux leaving the surface.. so u just some close gaussian surface enclosing the sphere

and then gauss laws says the net flus through that surface would be q/eps
so that must be your answer!

Answer 6

yea sig figs are WAY off :D :D!!

Answer 7

So... did I actually manage to do it correctly?

Answer 8

well.. you did :P

Answer 9

Awesome! :D
Thanks guys!

Answer 10

but what is important is whether you understood it or not!

Answer 11

Yeah, I am getting this stuff better and better :P

Answer 12

kudos!!! :D

Answer 13

Wait.... I just plugged in these answers and it says they are wrong.

Answer 14

Here we are.

Answer 15

Wait, first one I needed to multiply by 9.5E-6

Answer 16

(a) .000026
(b) \(2.93785\times10^{6}\)

Look better?

Answer 17

q IS NOT EQUAL TO 4 pi R square... lolzz

haha!! q = (density) times (4 pi R^2)

Answer 18

Both are correct now! :D

Answer 19

I forgot to multiply by density lmao.

Answer 20

@Mashy Yes. I was wrong. Flux is integral of E field times area.
Closed surface integral (Gauss) would = q/epsilonsub0. I will delete.

Answer 21

@douglaswinslowcooper yup .. just a silly mistake.. but not as silly as austinL dude's mistake :D :D!!

Answer 22

We are racing to the bottom!