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The Lorentz force experienced by a charge $q_1$ is: $$\mathbf F = q_1(\mathbf E + \mathbf v \times \mathbf B)$$ where x means vector-product. The electric field $\mathbf E$ between charges does not come from magnetic properties of the charges. With the magnetic field the things change. A moving charge $q_2$ produces a current, and a current produces ...

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$d$ can be infinite only in math books. Your statement about Coulomb's Law is true. if $F=0$ then $q_1q_2=0$. As $d$ gets arbitrarily large $F$ will get arbitrarily small. You can find a $d$ that makes $F$ as small as any number you choose ... except zero. I'm interpreting your question as being about Coulomb's Law itself, not any practical application ...

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What Jim is talking about in the comments is a limit: for $d$ becoming increasingly large, $F$ becomes increasingly small. For example, if $q_1 = q_2 = 1\,\text{C}$ and the charges are a distance $d = 1\,\text{m}$ apart, we find that the force is $$F = k \sim 10^{10}\,\text{N}.$$ If the distance is made 1000 times bigger, the force is F = ...

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It seems like you are trying to calculate the electric field due to a line charge distribution at a perpendicular distance z from mid-point of the rod. The given integral can be solved by taking x=z tan(theta). But to observe the physics behind it, in figure draw the triangle and observe where is this theta such that tan(theta)=x/z. Now instead of taking ...

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Hint: Put $x=z\tan \theta {}{}{}{}{}{}{}{}$

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The reason is that Coulomb's law is only directly valid for point charges, i.e. for charges, sizes of which are much smaller than distance between them. For particular symmetry reasons it appears to also be applicable to spherically symmetric balls of charge - but note that the distance you should take is not between the surfaces - it's between the centers ...

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