Recent content by Berko

Your ninth line of work...one of your minus signs should be a plus sign. Fix those and redo. I'll look at what you come up with.

Your seventh line of work...you squared the right side but not the left.

Anyone know where I can find a discourse on the dirac delta function in spherical or polar coordinates, in particular why it is the form it is with correction coefficients? Thank you.

From your last line, isolate the square root term and square both side...get rid of those square root signs.

I don't understand the issue. If .99... does not equal 1, there must be some number in between. Nothing is in between so .99... = 1. Why is this an issue?

Weeee!

yup. looks good. I'll do it quick here to make sure we both get the same answer. One sec.. I get 42 Joules (to 2 sig figs)

Why is it impossible? This is a non realistic problem that has no friction on the table. The block slides down, producing an acceleration of the ramp to the left, while the block accelerates to the right. Of course, if the block does not slide down, then there is friction between the block and...

So, here's what I came up with. Would love confirmation. The E field is generally sigma/2epsilon. Since it's a conductor, the E field is sigma/epsilon. So, the rest of the conductor must be supplying the needed E field to cancel the inside E field and reinforce the outside E field. How much...

I'm having a thought. Can it be due to the fact that the electric field (sigma/epsilon_nought) is due to the entire surface and we are only interested in the field the REST of the surface produces at the surface patch under consideration?

I loaded the pdf. It starts on page 894.

What page in the book? I don't se any pages labeled 17 and 18.

But, that's exactly my point. The E field SHOULD be sigma/epsilon_nought...but it IS sigma/2epsilon_nought.

On page 60 of his 3rd edition of Classical Electrodynamics, he discusses the method of images applied to a grounded conducting sphere with a single charge q outside it. Near the end of the problem, he calculates the force on a small patch of area da as (sigma^2/2epsilon_nought)da. Now, it...

Yes, you are correct. But, in general, "infinity" times zero is not 1...it can be anything, which is why we do not allow it.