Recent content by Laudator

Em... I'm only getting more confused... The problem in the book asked the total energy stored in the configuration, the answer is the one I mentioned, the author didn't take the central point charge into account. He stated that the first step of forming this configuration is to move in the...

I think I got the second part of my question worked out, thanks to @vanhees71, now I know the smearing of ##dq## takes no work. Here's my train of thought: I will treat the conductor as two conducting spheres with nothing in between. Originally, there's only a point charge at the origin and...

I suppose you meant "tangent to"?

@Andrew Mason @vanhees71 While calculating the energy stored in a charged sphere, I read from here. The author started from a sphere which is already charged with ##q## and then bring ##dq## from infinity onto the sphere with radius ##a## and the work it takes is:$$W=\int^a_\infty\vec F\cdot...

First, I applied Gauss' Law for ##r>b##, the total charge inside is ##(+q)_\text{point charge at O}+(-q)_\text{charge on sphere a}+(+q)_\text{charge on sphere b}=q##, so the field behaves like a point charge. The potential of a point charge is given by ##V=\frac{1}{4\pi\epsilon_0}\frac{q}{r}##...

I don't understand what you are suggesting.

It's ##+q## and according to Gauss' Law, the field behaves like a point charge when ##r<a##, so ##\mathbf{\vec{E}}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\mathbf{\vec{r}}##, that's what's confusing for me because ##\mathbf{\vec{E}}=-\nabla V##, adding a constant to ##V## won't change the field...

A conducting spherical shell with inner radius ##a## and outer radius ##b##, a point charge, ##+q##, is located at the center of the cavity, inducing ##-q## on the cavity wall and ##+q## on the shell. First, I want to know the potential in different region of this configuration. My thought...

Follow up: Found an answer on StackExchange says “... if the integral vanishes for all possible regions of integration, then the function is zero everywhere.", he also pointed out the formal proof of this is not easy. So I guess at my level, I should settle for this.

Stokes' Theorem states that: $$\int (\nabla \times \mathbf v) \cdot d \mathbf a = \oint \mathbf v \cdot d \mathbf l$$ Now, if for a specific situation, I can work out the RHS and it's equal to zero, does it necessarily mean that ##\nabla \times \mathbf v = 0##? I mean all that tells me is that...

@BvU ... How stupid am I ... Thanks ...

I know the area of a thin ring of radius ##r## can be expressed as ##2\pi rdr##, however, I wonder if I use the usual way of calculating area of a ring, can I reach the same conclusion? I got this: $$4\pi(r+dr)^2-4\pi r^2=4\pi r^2+8\pi rdr+4\pi (dr)^2-4\pi r^2=8\pi rdr+4\pi (dr)^2$$And now I'm...

@PeroK, thanks a lot, that's of great help.

@PeroK, thanks for the tip, I think I got it, here's my result, I hope I got it all worked out correctly: 3 points are labelled A(a,0,0) B(0,2a,0) C(0,0,a). Two vectors are then formed using point A,C and B,C, which gives: $$AC=a\vec x - a\vec z$$ $$BC=2a\vec y - a\vec z$$. The area of the...

@DrClaude, @PeroK, thanks for the info, I've already edited the main post, hope that's enough.