Question

@TuringTest

Answer 1

i'm having issues with proportionality.

Answer 2

ok here is a question

Answer 3

Two conducting spheres, one having twice the diameter of the other, are separated by a distance that is large compared to their diameters. The smaller sphere has charge q and the large sphere is uncharged. They're connected by a long thin wire.
The charges on the smaller and large spheres, respectively are....

Answer 4

I should probably start by considering their potentials.
\[V_1=\frac{kQ}r\]
\[V_2=\frac{kQ}{2r}\]

Answer 5

These potentials are equal correct?

Answer 6

they are connected?

Answer 7

yes, by a long thin wire

Answer 8

then the potentials of both must be equal, or there would be a current in the wire

Answer 9

oh ok,
so...
\[\frac{kQ_1}r=\frac{kQ_2}{2r}\]

Answer 10

right

Answer 11

uhm.....

Answer 12

am I solving for Q_1 and then Q_2?

Answer 13

what are you trying to show, the proportion of the surface charges? if so, you need to write Q in terms of \(\sigma A\)

Answer 14

Uhm let's see...it's MC and the options are
\(\frac Q 2\) and \(\frac Q 2\)
\(\frac Q 3\) and \(\frac {2Q} 3\)
\(\frac {2Q} 3\) and \(\frac Q 3\)
and so on....

are they looking for some sort of proportion?
This is a practice MC question

Answer 15

"the charges on the smaller and larger sphere are respectively:"

Answer 16

the way I would do it is to get the relation of the surface charge densities, then you can compare the surface charges

Answer 17

\[\sigma=\frac{Q}{4\pi r^3}\]?

Answer 18

oops r^2

Answer 19

\[\sigma_1=\frac{Q}{4\pi r^2}\]
\[\sigma_2=\frac{Q}{4\pi(2r)^2}\]

Answer 20

um... not how I did it, but it should get the right answer
you ought to differentiate between Q1 and Q2 though

Answer 21

so the charge densities are different, but the voltages are the same.....where from here? Electric fields?

Answer 22

remember our derivation we did once?

Answer 23

oh yes sorry...the Q's are different
\[\sigma_1=\frac{Q_1}{4\pi r^2}\]
\[\sigma_2=\frac{Q_2}{4\pi(2r)^2}\]

Answer 24

\[~~~~~V_1=V_2\]\[\frac1{4\pi\epsilon_0}\frac{Q_1}r_1=\frac1{4\pi\epsilon_0}\frac{Q_2}r_2\]\[\frac1{4\pi\epsilon_0}\frac{A_1\sigma_1}r_1=\frac1{4\pi\epsilon_0}\frac{A_2\sigma_2}r_2\]\[~~~~~~~~~~~~~~~~~\vdots\]

Answer 25

ohhhh....yes

Answer 26

\[\frac1{4\pi\epsilon_0}\frac{A_1\sigma_1}r_1=\frac1{4\pi\epsilon_0}\frac{A_2\sigma_2}r_2\]
\[\frac{A_1\frac{Q_1}{4\pi r^2}}r=\frac{A_2\frac{Q_2}{4\pi(2r)^2}}{2r}\]
\[\frac{A_1{Q_1}}{4\pi r}=\frac{A_2Q_2}{2r}\]
now I can cross out the r's?

Answer 27

should I have written out the definitions of A?

Answer 28

and I forgot a pi an the other side

Answer 29

sub in for the A's first so you can see where they go

Answer 30

ok

Answer 31

yeah you should write it out, because the radii matter

Answer 32

\[\frac{4\pi r^2{Q_1}}{4\pi r}=\frac{4\pi (2r)^2Q_2}{4\pi2r}\]
\[\frac{r{Q_1}}{1}=\frac{ 2rQ_2}{1}\]
I must have made a mistake....gosh

Answer 33

It's just algebra from here on correct? I think I can just check it over...

Answer 34

the Q's should be sigma's for one thing

Answer 35

actually that's the only mistake you made

Answer 36

but don't I need the answer in terms of Q's?

Answer 37

we'll get that back easily once we have the relationship of the densities
there may have been another way to do it that did not require us to find the sigma relationship, but I didn't see it.

Answer 38

oh ok