Recent content by kuruman

To determine the direction of the gravitational field, also known as "down", you suspend a mass from a string and you deduce that "down" is the direction from the point of support to the attached mass at equilibrium. This device is also known as a plumb-bob. The picture below is modified from...

Yes, indeed.

I am not so sure about that. The charge distributions on all surfaces can be sorted out by using Gaussian pillboxes. Refer to the figure on the right. 1. Construct Gaussian pillbox 1 enclosing all three plates (black). The total enclosed charge must be zero to eliminate the electrostatic...

Please provide a drawing and a concise statement of the problem. Also according to our rules, to receive help, you need to show some credible effort towards answering the question. How about telling us what you do know and how you would approach this problem? You need to be more specific about...

Why? If the plate is truly infinite, it must be at the reference value of the potential at infinity, i.e. earthed, at all times. If you place charge Q on one side, it will leak out to infinity. I think that my concentric shell picture (I wouldn't call it a model) gets around that problem and...

Here is my resolution. Consider two concentric conducting shells. Charge ##Q## is placed on the inner shell and then the outer shell is earthed. The inner shell has outer radius ##R## and the gap between shells is ##d##. What are the surface charge distributions on the shells? (See figure on...

Look at the figure on the right. The total force holding the right half in place is ##F_{\text{total}}=2pA## where ##A## is some appropriate cross sectional area. Can you provide a derivation or reference for ##p=\frac{B^2}{2\mu_0}~##?

Yes, however the connection between the explanation in post #11 and the charges shown in the figure is not, in my opinion, made rigorously. I am wondering whether the OP will be able to use his approach and arrive at the correct solution if the initial charges on the plates, before earthing, are...

Look at the figure on the right. It shows plates 3 and 4 and the standard pillbox Gaussian surface shown as a dotted line. There are surface charge densities as shown. Using Gauss's law is a better idea than using work done because it relates the surface charge density directly to the electric...

This reformulation makes sense.

Then the statement of the problem "4 conducting plates with charges Q, -2Q, 4Q and 3Q are present as shown in the image attached. The third plate is grounded." is misleading. The attached image shows plate 3 already grounded and the total charge on each plate at that time. Is it possible that...

Why? This means that the electric field on the left side of the plate array must be zero. If you construct a Gaussian pillbox that includes all four plates (see below), you see that the net charge enclosed is +6Q which, according to Gauss's law, means that there is a net electric field...

Not quite. If you look in the figure, they write ##\Delta m =-|\Delta m|.## This means that ##\Delta m## is negative. That is also explicitly mentioned in the text I think you are confusing a symbol representing an algebraic quantity with its value. When you write ##x+2=0##, you can see that...

Furthermore, what matters in variable mass problems is the rate of change ##\Delta m/\Delta t##. Since ##\Delta t## is always positive, when the system loses mass (as in the case of a rocket) the ratio is negative; if it gains mass (as in the case of a moving container accumulating rainwater)...

Your ideas are correct and you can take it from there. However, since you asked, you can do it more directly by noting that there are ##N## turns distributed over ##\frac{\pi}{3}## so that $$dN=\left(\frac{N}{\pi/3}\right)d\theta=\frac{3N}{\pi}d\theta.$$Since the variable of integration is...