# Tag Info

## Hot answers tagged electrostatics

4

As Mostafa says, it is macroscopically at equilibrium, not necessarily microscopically. There may be one misunderstanding you have, which is about "surface". I will talk about it later. In my opinion, equilibrium should be understood as no electron moving. It is easily to show that the electric field in conductor is zero. If the electric field is non-zero, ...

2

You simplified incorrectly, the result is zero. I assume you computed it using floating point arithmetic, which explains why the discrepancy is on the order of machine epsilon. Try using symbolic manipulation in Mathematica: Sum[Cos[2 \[Pi] n/13], {n, 13}] // FullSimplify Out: 0

2

When an electric dipole is placed in a uniform electric field making an angle with the direction of the field as shown in the figure. Force on charge $-q=-q\overrightarrow{E}$ (opposite to $\overrightarrow{E}$) Force on charge $+q=q\overrightarrow{E}$ (along $\overrightarrow{E}$) Thus, electric dipole is under the action of two equal and unlike ...

2

It seems to me that you have more of a conceptual issue than a mathematical one. To hopefully remedy this, let me remind you of a couple of facts. Given an electric field $\mathbf E$, an electric potential $V$ for $\mathbf E$ is any scalar function $V$ for which \begin{align} \mathbf E = -\nabla V \end{align} It follows that if $V$ is such a potential, ...

1

I've posted an answer describing the derivation of potential energy which you might want to read, as the same argument applies to electrical potential and I think that's what you're missing. Basically, given an electric field, the first step in finding the electrical potential is to pick a point $\vec{x}_0$ to have $V(\vec{x}_0) = 0$. Then, to determine the ...

1

$\def\l{\left}\def\r{\right}$ You can work in the complex plane. The real component is your $\hat i$-component the imaginary component is your $\hat j$-component. There $\l(\cos(\phi),\sin(\phi)\r)$ is represented as $\exp(i\phi)$ with $i=\sqrt{-1}$. In this context your sum is $$\sum_{k=0}^{N-1} e^{i2\pi\frac kN}.$$ if we substitute \$a := ...

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