Recent content by Steve4Physics

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    Cone on horizontal surface

    The non-slipping condition is simply required to enable the cone to rotate around the vertical axis. Friction provides the centripetal force. On a frictionless surface, the centre of mass would only move in a straight line.
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    Cone on horizontal surface

    That's a pity - I thought it [my Post #10 formulation] was quite good! How could it be improved? Yes - then the solution is nearly complete (as already hinted-at by @haruspex). The question does not involve a non-slippery condition. Or maybe you are referring to @Lnewqban's additional (and...
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    Cone on horizontal surface

    Possibly something like this... A solid right circular cone of uniform density has mass m, height h and a base of radius r. ‘V’ is the cone's vertex. If required, use the fact the cone’s centre of mass is located on the cone’s internal axis, a distance 3h/4 from V. The cone rolls, on its side...
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    Curving of gravitational field lines

    A line can't be perpendicular to a point. You probably meant to ask: "Aren't gravitational field lines always straight and directed towards a body's centre of mass?". To add to what’s already been said, even single objects can have curved gravitational field lines. E.g. the shape of the earth...
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    Earthed plates confusion

    I'll explain! In Post #24 you wrote: I looked at the link. The first (remarkable) sentence of the abstract is this: "The foremost grail of this academic indagation is to delineate a mathematical expression of the normalised charge density over a flat disk." Had to look-up 'indagation' and...
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    Earthed plates confusion

    My wording was very poor. It was not the intended meaning. (But I hope the intended meaning was clear from the latter part of the post.) I’ve updated the wording. Caution needed. Net zero charge is not guaranteed. As already noted by @SredniVashtar (Post #14) and @haruspex (Post #16) it's...
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    Earthed plates confusion

    I may misunderstand the issue you are indigating. But to try to answer your question... Take three conducting plates (equal sizes, aligned, parallel, negligible gap compared to linear dimensions). One outer plate is grounded and the other two plates have charges of, say, +5Q and +2Q. I believe...
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    Earthed plates confusion

    Well, I'm not sure if this answers your question but here’s a qualitative (definitely not rigorous) attempt... Thinking visually, a field line starts on a positive charge and ends on a negative charge. Consider two (finite) parallel conducting plates arranged as shown below. The upper plate is...
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    Earthed plates confusion

    @tellmesomething, this might resolve specific issues you raised…. ‘Ground’ and ‘infinity’ are not necessarily equivalent. I believe that we can only say that no work is done bringing a test charge from ground to the right side (or indeed to the left side) of plate 3. Minor (pedantic) point -...
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    Earthed plates confusion

    Hi @tellmesomething. I understand what you mean - sorry for not picking it up sooner. I’ve edited what I said in Post #8 as it may be incorrect/misleading. I too had assumed the charges were distributed as you show in your Post #11 diagram. Maybe this is the problem. Or maybe there’s no...
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    Earthed plates confusion

    The initial net charge is non-zero. But when plate-3 is grounded, charge will flow between the system and ground. The final net charge will be zero and the outer surfaces will have zero charge. I think this result is quite general for conducting/earthed systems – providing there is no external...
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    Earthed plates confusion

    The Post #1 diagram is misleading. It would be better to include an open switch in the earthwire and ask something like: "What are the final charges on each surface of the plate after the switch is closed?". Edit. Aha. @kuruman beat me to it..
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    Two-layer dielectric, four-capacitor model vs three-capacitor model

    I don’t think that’s correct – there will be discontinuities in charge-densities and field-strengths as you move left to right. To add what @mjc123 said, I think the system should be visualised like this:
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    Chain falling out of a horizontal tube onto a table

    FWIW, on the right-hand side you are adding quantities with different dimensions: ##v## has dimensions ##LT^{-1}##, ## \frac {dv}{dt}## has dimensions ##LT^{-2}## and ##vgx## has dimensions ##L^3 T^{-3}##.
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    Chain falling out of a horizontal tube onto a table

    @NTesla, maybe this is one source of your difficulty... That's not entirely correct. The horizontal section of the chain is accelerated by tension,T, acting at its right-most end: We have to treat the falling vertical chain section as if there are only two forces acting on it*. The forces...