Question

A parallel plate capacitor has a capacitance C and charge Q.Another plate is inserted in between the plates of the capacitor at the center and is connected by a long wire to far off spherical conductor of radius R. What is the potential difference between the edge of the sphere and one of the outer plates?

Answer 1

@agent0smith @rvc

Answer 2

@ganeshie8 @hartnn

Answer 3

my answer is not matching..

Answer 4

I have a doubt.. why have they mentioned that the spherical capacitor is "far off" ? Is it of any significance..?

Answer 5

@Michele_Laino @Vincent-Lyon.Fr @IrishBoy123

Answer 6

@mayankdevnani

Answer 7

is it to specify that the spherical capacitor is isolated ?

Answer 8

is the answer Q/C ?

Answer 9

@mayankdevnani no..that's not the answer..

Answer 10

great !

Answer 11

let me try this again and get you back soon !

Answer 12

okay

Answer 13

@samigupta8

Answer 14

V=CQ/R

Answer 15

Plate inserted between the parallel plates of capacitor ..
Ryt?

Answer 16

Since you assume the electric field to be homogeneous between the two fields, it comed out to just E = V / d, where V is the voltage between the two plates and d is the distance between them. This field is cause by some constant charge density ρ ( ρ = Q / A, where A is the area of the plate) - in this case E = ρ / ε.

The voltage is the line integral of the electric field between one of the plates and the inserted one. \[V = \int\limits_0^{\frac{d}{2}} E \;dz\] Assuming your z-axis perpendicular to the plates and one of them is at z=0. This integral is completely trivial and you just need to plug in what we stated before.

Answer 17

yes @samigupta8 .. then?

Answer 18

You're trying to find V, so substituting E for V / d will not get you anywhere, will it? :)

Answer 19

no.. i didn't mean like that.... i got it thanks !

Answer 20

Did u get it?? How??

Answer 21

Q/2C is the answer!

Answer 22

@MuH4hA 's method is correct.. simply put it is V= Q/ C' where C' is the new capacitance.. which is 2C since d has reduced by half.. so V= Q/2C.. that's it..!

Answer 23

got it? @samigupta8

Answer 24

Okk bt priyar v have to like calculate the potential diff between the outer plate and sphere ..
So don't you think that capacitance will be like the series combination of three capacitor ..

Answer 25

Like first capacitor is in between first plate(lhs) and the other plate's lhs
Then between the other plate's ..
And the last one between the other plate's (rhs) and the first plate (rhs)

Answer 26

Yes - you can just make a symmetry argument and be done with it. However, this only works for an ideal plate-capacitor. (large plate surface in relation to d, so we can ignore the inhomogeneities at the edges and a constant ε). Actually solving the integral would work for other geometries as well - but it might be quite hard.

Answer 27

i will tell you later

Answer 28

@muh4ha will u plss explain clearly?

Answer 29

@samigupta8 Explain the symmetry argument? Or the integral? What seems to be the problem? :)

Answer 30

Well! I m not linking anything with your way...
I just wanna make sure if we can solve for the three capacitance and solve fof the net capacitance and get the value of charge

Answer 31

For*

Answer 32

I'm not sure I follow (what's the "three capacitance" ?).. are you trying to think of it as 2 capacitors in series?

\[\frac{1}{C_{total}} = \frac{V}{Q} = \frac{V_1 + V_2}{Q} = \frac{V_1}{Q} + \frac{V_2}{Q} = \frac{1}{C_1} + \frac{1}{C_2}\] or more general for N capacitors in series: \[\frac{1}{C_{total}} = \sum_{n = 1}^N \frac{1}{C_n}\] (like N parallel resistors).

If you think about two connected capacitors in series, in the middle between them will be two plates (one of each capacitor) connected by a conductor. These two plates are physically disconnected from the rest of your electric circuit and thus the charge on them must remain constant. If we assume there is 0 charge on them, when no voltage is applied over the whole of the 2 capacitors, there'll still be 0 charge when some voltage is there. Therefore, the two connected plates in the middle must have either 0, or equal and opposite charge +/-Q from each other (to cancel out). Then, if you look at the capacitors on their own, since one plate will have the charge +/- Q, the other plate must have the opposite charge and you end up with both capacitors carrying the same charge Q. The voltage drops are of course different in general, but the total drop will be the sum of the two (since they are in series) and we can write V/Q = (V1 + V2) / Q.

In our example, the two capacitors are identical (plate is inserted exaclty in the middle), therefore they will have to have the same capacitance, same stored charge and there'll be the same voltage drop on the individual capacitors.

Answer 33

here is my reasoning:
the situation, is like below: