Question

Two conducting spheres P & Q have radii 'R' & 'r' respectively. They are connected by a thin wire, charged and then separated to a large distance.
Now at the surface of P & Q, the ratio of electric fields is: -

A.) \(\Large{r/R}\)
B.) \(\Large{R^2/r^2}\)
C.) \(\Large{r^2/R^2}\)
D.) None of these.

Answer 1

@experimentX plz help:)

Answer 2

@Rohangrr @ajprincess @UnkleRhaukus Plz help :)

Answer 3

none of these

Answer 4

but in my book answer is A.) r/R

Answer 5

because it says " P & Q have radii 'R' & 'r' respectively. They are connected by a thin wire, charged and then separated to a large distance.
Now at the surface of P & Q, the ratio of electric fields is"
it should be

Answer 6

but my book's answer is different:\

Answer 7

it is option A.

Answer 8

correct I didnt look the last bit of it

Answer 9

when two charges spheres are connected by a thin wire ... the whole potential of both spheres must be equal.

Answer 10

@maheshmeghwal9 follow this : http://www.physics.purdue.edu/academic_programs/courses/phys241/exam_archive/Ex1_F02_with_solutions.pdf

Answer 11

i haven't studied potential yet:(

Answer 12

is there any solution to this question based on gauss's law?

Answer 13

becoz i have studied
\[E*A=\frac{q}{\epsilon_0}.\]

Answer 14

or any other solution without the concept of electric potential?

Answer 15

thinking

Answer 16

ok:)

Answer 17

I have found this much
\[E_1=\frac{q}{4 \pi R^2 \epsilon_0}\]
\[E_2=\frac{q}{4 \pi r^2 \epsilon_0}\]
\[\frac{E_1}{E_2}=\frac{r^2}{R^2}.\]q = charge gained by P = charge lost by Q
but this is wrong; i know:(

Answer 18

probably they have charge distribution according to their sizes

Answer 19

may be.

Answer 20

since it's sphere ... it must be symmetric.
\[ E_2=\frac{(r/R)q}{4 \pi r^2 \epsilon_0}\]
don't put those 4 pi r^2 unless it's necessary

Answer 21

sorry i don't get u
how u gt E_2 as above?

Answer 22

i mean from where did this r/R come?

Answer 23

i told you .. both don't have same charges.

Answer 24

but if they are connected through a wire they must gain or lose equal amount of charges
i think

Answer 25

R has Rq charge and
r as rq charge.
depending on radii

Answer 26

if R as q charge then r as r/R q charge.

Answer 27

going offline ... i'll won't be back in less than 2 hrs

Answer 28

np:)

Answer 29

r/R makes perfect sense because the surface area of a sphere is:\[SA_{sphere}=4\pi r^2\]Conducting spheres will hold charges proportional to their surface area. You don't need Gauss' Law for this.

Answer 30

but how to get r/R
mathematically?

Answer 31

\[\frac{4pir^2}{4piR^2} \propto \frac{r}{R}\]

Answer 32

Actually, I guess you'd really get:\[\frac{4pir^2}{4piR^2} = \frac{r^2}{R^2}\]

Answer 33

really confusing:\

Answer 34

The more I think about it the more I convince myself that regardless of what the book says, the answer should be r^2/R^2 :/

Answer 35

i also thought this first time as in my solution:)
i think we should wait for others also to confirm.

Answer 36

Yes :)

Answer 37

:)

Answer 38

I found on an old exam online which almost makes sense:

Since the spheres are connected by a conducting wire, they are initially at the same potential:\[k\frac{q_1}{r_1} = k\frac{q_2}{r_2} => \cancel{k}\frac{q_1}{q_2} = \cancel{k}\frac{r_1}{r_2}\]The electric field at the surface of each shell is: \[k\frac{q}{r^2}\]Thus the ratio of electric fields is:\[\frac{E_1}{E_2} = (\frac{q_1}{q_2})(\frac{r_2^2}{r_1^2})\]From the previous equality:\[\frac{E_1}{E_2} = \frac{r_2}{r_1}\]

Answer 39

thank u very much:D

Answer 40

Maybe I'm just not awake yet but I'm still not seeing how we go from r^2/R^2 to r/R....unless the spheres are assumed to have the same radius in which case the question would not have listed r and R.