Recent content by Guest812

About 1,000 years ago, a more accurate calculation of the Earth's radius was made by Al-biruni using a protractor, tape measure, and trigonometry as shown in the below video:

How to measure the Earth's radius using a watch, tape measure, and trigonometry: It is possible to easily see the effects of the Earth's curvature by simply observing the Sun set while you lie on the ground, and afterwards see the sun set again approximately 10 seconds later if you stand up...

OK guys, thanks for setting me straight. I drew a uniform E-field, then drew a nonconductive object inside the field, then electrostatically induced "polarized" charges on opposite sides of the object, and after looking at the diagram, agree no net movement of the complete object, just...

Then please explain how a charged object (i.e. something that produces an E-field) can attract any (i.e. insulating or conductive) uncharged object? In the meantime, I'll try to find the chapter in the free MIT text that explains the phenomenon better than I did.

An uncharged insulated balloon (or anything else: even a conductor) will experience an inductive electrostatic force (i.e. if lite enough will accelerate and move) if inserted into any E-field. The reason anything will experience an electrostatic force if inserted into any E-field is because...

NO ! A sphere only appears as a flat surface when you're extremely close to the sphere's surface. The further away from the sphere you are, the more the sphere will appear as a point charge. In other words, when viewed extremely close to the conducting sphere's surface, the E-field lines...

Good. Then from there, the only way I know from the top of my head to intuitively view what happens inside and outside a conducting sphere, is visualize how the E-field behaves when it is measured infinitely close to both sides of the sphere's surface, where infinitely close to a sphere's...

The text I change to red is incorrect, in the context you used it in. The correct and more accurate phrasing is: "an infinitely long flat conducting plane will have a constant E-field, regardless of how far away from the plane the E-field is measured. Note an infinitely long flat conducting...

Yes. In my above posts, I diagrammatically explained it using a thick flat conducting plate with equal surface charge on both opposite sides analogy, then explained how a conducting sphere of infinite diameter could be treated as a conducting flat plate, and how even the tiniest conducting...

If the sphere diameter is infinitely large, near the infinitely large sphere's surface, the sphere's E-fields will behave as if the sphere's surface was flat. A tiny sphere, can treated as an infinitely large sphere, if you measure extremely close to the sphere's surface. If you know how the...

The excess charges do experience a force. The charges force each other away from each other, like a ballon's wall is forced outward when you add more "gas charge" inside the balloon. The main difference between how an excess "gas charge" inside a balloon behaves and an excess electrical charge...

No, for a very long flat conducting plate (regardless of how thick the plate is), the E-fields will be constant no matter how far perpendicular away you from the surface, regardless of whether you're measuring from the charged surface's left or right side. I recommend you study how to calculate...

If excess charges are placed on a very long thick conducting flat plate, the excess charges will redistribute themselves until their sideways (tangent to the surface) forces are equal and opposite to each other. When the all the charges on a conducting plate are redistributed so all their...

I've seen all the above mentioned books and think they all look good. Except for the Halliday & Resnick book I read during college, I can't comment on which is best, because I haven't completely read all the others. One E&M "textbook" which hasn't been mentioned yet is the one available for...

Thanks Using the approximation you recommended, I was able to continue where you left off, as follows: d ln(n) + ln(1+\frac{1}{n}) + ... + ln(1+\frac{d}{n}) d ln(n) + \frac{1}{n} + \frac{2}{n} + ... \frac{d}{n} d ln(n) + \frac{1}{n} \cdot (1 + 2 + ... d) d ln(n) + \frac{1}{n} \cdot...