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If you are talking about point charges then, as explained above, the answer is no. But in the case of non-uniform charge distributions, it is possible for same-charge particles to attract, if they are sufficiently close. As an example, the following two particles are identical, each having a net charge of -1. Plotted below them is their potential energy as ...


2

You have two questions, and they have different answers. First of all, let's be clear about what Gauss's law is in integral form: $$ \int \vec{E} \cdot \mathrm{d} \vec{A} = \frac{Q_\mathrm{encl}}{\epsilon_0} $$ In words: the total flux integrated over a closed surface is equal to the charge enclosed in that surface, divided by the permittivity of free ...


1

This is best understood by approximating the dipole as a pair of finite charges $\pm q$ separated by a finite distance $d$. In a uniform electric field, the electrostatic forces on each of the charges will cancel out exactly, but in a non-uniform one the forces on the two will be slightly different, leading to a slight imbalance and therefore a non-zero net ...


1

The electric field is a vector quantity, representing the electric force per unit charge acting on a test particle at a particular position in space. Since force is a vector, the electric field too is a vector quantity. The electric potential however is not a vector. The electric potential is the amount of electric potential energy that a unitary point ...


1

Well, if you have an extremum (say local maximum) of $W$ at $p$, then you have a small open ball $N$ centered in $p$ such as $W(x)< W(p)$ for some $x\ne p$ in $N$. Therefore $$ \frac{1}{{\rm vol}(N)}\int_N W(x)dx < W(p)~~.\tag{1}$$ assuming we have taken the ball $N$ small enough for $W(x)<W(p)$ for some $x\ne p$ in $N$. But if $\nabla^2W=0$, by ...


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Laplace's equations, as well as all other kinds of fields equations, always describe a local property, whether or not the right hand side is non-zero. In particular, it restricts all the possible fields configurations to the ones obeying $$ \nabla^2 \phi(x,t)= f(x,t). $$ It describes the local properties of the field $\phi$ upon the space in the region where ...


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Ok, here is an answer in some simple words. Harmonic functions obey the following property: if you draw a circle around any point $x_0$, and take the average of the function over that circle, that average will be equal to the value of the function at $x_0$. Now, if $W$ is not constant, it must have a maximum somewhere. Let's call this maximum point $x_0$. ...


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The E field inside the conducting wire is 0, so what is it really doing? The potential difference between two points is related to the electric field along the path between them: $$V_{ba}=\int_a^b \vec{E}\cdot{}\vec{dl}$$ So the fact that the high-conductivity material forces a (near-)zero electric field is exactly why the two ends of the conductor ...


1

Think of a gas. We will ignore gravitational potential, and we'll consider the situation to be at steady state. Compressing the gas takes energy, so we conclude that the entire container of the gas will be at a constant level of compression - in this case, constant density as well. You may intuitively understand that this is true no matter how oblong the ...



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