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The traditional least-squares fitting or chi-squared minimisation route of fitting a straight line makes the implicit assumption that the errors on the x-axis quantity are negligible. If that is so, then there is no reason why you can't use the uncertainty in the gradient as the uncertainty in $R$. I guess from your question though, that this is not the ...


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The voltage V that you're computing would be induced between the upper and lower sides of your orange block, and $L$ in your computation is the distance between these two sides. This voltage is induced between any two opposite points on the upper and lower sides provided they are in the area that is covered by the magnetic field. If you connect a wire with ...


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Well, yes you can, but it is usually very hard. Here are the steps: Solve the Laplace equation: $$ \nabla^2V = 0 \, .$$ In your case, find the general solution in spherical coordinates. Try to use every simplification you can. You might wonder why you don't solve Poisson's equation: $$ \epsilon\nabla^2V = \rho \, .$$ That's because a conductor is an equal ...


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No. The capacitors are in series. This is because one side is at the same potential. But it looks like parallel. If you apply Kirchoff's Loop Law, you will see that they are in series. And in that way too, you will get that answer. Hope that helps.


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NO. The Capacitors are in series as when we go from one capacitor to another we find no junction in between.


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for same voltage supply, the power consumed by two resistances in series connection is less in compare to power consumed by same resistances in parallel connection. Therefore we can say that - P(series) < p(parallel)


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If you keep the resistance constant, then $V=IR$ means that voltage is directly proportional to current. If you keep the power constant, then $V=\frac{P}{I}$ means that voltage is inversely proportional to current. However, because $V=IR$, we can write that $P=I^2R$. Therefore, if we say resistance is constant, then power must change with current, which ...


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A metal conducts quite well because the there is an electron band that crosses the Fermi level. So, electrons can easily be excited to increase their momentum a bit and consequently move in one direction. Now if you add one electron to the wire, the Fermi level rises. However, you would not be able to see the increase caused by one single electron (or a ...


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Think about what exactly the rate of flow of electrons is? It is the number of electrons per second passing through! The number of electrons is not included in your expression, and that's the problem. Let start over but this time with the number of electrons $n$ included: $$Q=It \Leftrightarrow \\ en=It \Leftrightarrow\\ \frac{n}{t}=\frac{I}{e} ...


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You have to think more carefully about what exactly $Q$,$I$ and $t$ signify: In $$ Q = I t$$ $Q$ is the charge that is transported by the current $I$ during the time $t$. If you now write $$ \frac{I}{Q} = \frac{1}{t}$$ then this gives how many times a charge of $Q$ is transported by $I$ during one unit of time (second), since $t$ is the time to transport ...



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