The answer given is D. How do I do this problem? I find it very confusing.I gathered that potential of all three must be made the same and field experienced by both smaller conductors is zero, but I don't know what exactly to do here. How should I approach this problem?
@Hema I'm not sure how you'd approach this formally, but charge always goes to the outside of a conductor. So all the charge will be on the outer surface of the hollow metallic conductor C.
@Jasmine If you bisect a lens but leave the two halves pressed together it doesn't change the focal length. If you bisect the lens and move the two halves apart, i.e. put a distance between them, then it does affect the focal length.
OK, you've got several things going on there. Firstly you have the complication that one side of the lens is in water, and secondly the presence of the mirror.
I would start by removing the mirror and finding the position of the image. Are you OK with that step? Do you know how to calculate the focal length when one side of a lens is in a different medium?
@Jasmine then that answers the question doesn't it? Or are you considering the second image formed when the light reflects off the mirror and passes back through the lens?
OK. The image formed in the water now acts as an object. Suppose the image is formed a distance $d$ from the mirror. The light travels a distance $d$ to the mirror then reflects and travels a further distance $0.8m$ back to the lens.
So in effect we have an object on the right side of the lens at a distance $u=d+0.8m$
If the image was formed behind the mirror then for the plane mirror will the subsequent image after refraction act as virtual object for the mirror?
If the image was formed behind the mirror then for the plane mirror will the subsequent image after refraction act as virtual object for the mirror? @JohnRennie
@Jasmine Yes, although since this is a plane mirror the situation is simple.
If the image forms a distance $d$ behind the mirror (to the right of the mirror on your diagram) the mirror just reflects it so the real image is formed a distance $d$ to the left of the mirror.
The field inside a conductor is zero, but we're not considering the field inside a conductor. We're considering the field in a cavity in a conductor when that cavity contains some charge.
:46848032 yes, though I'm about to start work again for another 30 minutes or so. I'll be done around 08:00 UK time, which I think is 12:30 your time.
@Hema the induced charge is not distributed evenly over the surface of the cavity. The charge density at A will be greater than the charge density at B.
Working exactly at the surface is tricky, because then you have to wonder if charges at the surface should be treated as inside or outside the conductor. In general we can't define the field exactly at the surface.
@JohnRennie would you please tell me if my understanding is correct? If the red surfaces S1 and S2 are spherical gaussian surfaces, S2 has net flux through it zero as net charge enclosed is zero. S1 however contains only q and therefore has a net flux. This implies that field within the cavity is not necessarily zero?
And therefore field near A is not equal to field near B
We know here is a charge q inside S1, so we know there is a net field through S1. That's fine. Note that because the system isn't spherically symmetry we can't easily calculate the field at S1, but we know it will be non-zero.
For S2, we don't know the charge inside S2 is zero unless we assume that the induced charge is equal to -q.
But, we know the field inside a conductor is zero, so we know the field at S2 must be zero because S2 is inside a conductor. That means the total charge inside S2 must be zero.
But the argument is kind of the other way round to what you said. We start with the fact the field inside a conductor is zero, and from that we infer the induced charge.
Imagine drawing a gaussian surface (a sphere) around the two charges. The desnity of the field lines is greater on the right hand charge, so when you integrate over the surface you'll get a larger number. Since the integral is $q/\epsilon_0$ that means the right hand side must have a larger charge.
But there is symmetry..... The electric field In every quadrant is same so finding electric field in x direction of a quadrant and multiplying by 4 should do the trick?
But there is symmetry..... The electric field In every quadrant is same so finding electric field in x direction of a quadrant and multiplying by 4 should do the trick?
The charge distribution is symmetric about the X axis and antisymmetric about the Y axis, so the field is always going to have a zero Y component and a non-zero X component. So the answer is either (a) or (b).
In case of closed surfaces the area vector is directed outwards the surface. But what is the direction of the area vector in case of an open surface e.g. A thin lamina type of surface. Does it depend upon the curvature of the surface i.e. concave or convex side.
In case of closed surfaces the area vector is directed outwards the surface. But what is the direction of the area vector in case of an open surface e.g. A thin lamina type of surface. Does it depend upon the curvature of the surface i.e. concave or convex side.
But the dot product is bracketed so it has to be evaluated first. The three dot products evaluate to $|a|\cos\theta_{ad}$, $|b|\cos\theta_{bd}$ and $|c|\cos\theta_{cd}$ so it's not obvious what common factor you'd remove.
In case of closed surfaces the area vector is directed outwards the surface. But what is the direction of the area vector in case of an open surface e.g. A thin lamina type of surface. Does it depend upon the curvature of the surface i.e. concave or convex side.
