6

It is unlikely that the electronics of the device is responsible for that. The times of high static voltage in TVs are long gone. The only electronic thing I think could be responsible for that is improper grounding. But I think even a manufacturer of cheap TVs cares about proper grounding, unless he wants to risk expensive lawsuits filed by the relatives of ...


3

This is a common misconception. A dipole is NOT a set of two opposite charges, so you do not have a dipole in your configuration. To be precise, you have a "physical dipole", two equal (in absolute value) and opposite ( sign) charges near each other. This is a usual configuration; but, as you have a finite numebr of point charges, you find the ...


2

UPDATE: Since the other answers are providing more of the answer, let me reveal my interactive visualization. As I mentioned in the comment to the OP, the "physical dipole" (composed of two equal-magnitude oppositely-signed point charges) is not the same as the "pure [or point] dipole". For the same dipole moment $\vec p$, they disagree, ...


2

Your teacher is correct. The force on one charged particle due to another does not change. But the medium (due to polarization) changes the net force on the particles. When the medium becomes polarized due to the presence of the charges, the electric field on each particle will be different compared to the case of the charges in a vacuum, meaning the net ...


2

The introduction of the seemingly finite volume $V$ obscures the fact that you are actually just introducing a constant charge distribution $$\rho_0:=\frac{N}{V}=const.$$ over the whole $R^3$-space, and this has of course infinite energy because it represents infinite charge. Neither $V$ nor the Yukawa potential is of any relevance here. Instead of using a ...


1

In simple terms: think about two charges A and B in vacuum, the Coulomb force acts on each charge. If the medium is changed, taking for example dielectric, the electric field of the charges change the orientation of the dipoles of the medium. It means that the medium is polarized. Polarization is the average dipole per volume. Then the medium itself create ...


1

I think I see my logical flaw. $∮E⃗ ⋅da=0$, however this does not mean that there is no E-Field passing through, right? It only means that the FLUX is $0$. So Gauss' Law holds, but we must be careful. I'll leave this up nonetheless, in case anyone comes against the same question again.


1

First, you should compute the total charge difference by use of one of the following methods: The definition of electric potential for a conducting sphere with radius $R$, i.e. $V = \frac{q}{{4\pi {\varepsilon _0}R}}$: This yields $\Delta q = 4\pi {\varepsilon _0}R\Delta V$. or The definition of capacitance of a sphere, i.e., $C={4\pi {\varepsilon _0}R}$ ...


1

The potential near the two charges would be found by adding two terms. The formula you show is a good approximation if, R, is much larger than, a.


1

The electric field near the surface: The sheet charge density on the spherical surface $$ \sigma = \frac{Q}{4\pi r^2}. $$ This renders an electric field $$ E = \frac{\sigma}{\epsilon_o} = \frac{Q}{4\pi \epsilon_o r^2}$$ The field from considering the surface charge density is exactly the same as that using Gauss's law. Induced charge dipolar layer: Yes. You ...


1

Consider the problem of determining the potential outside a grounded conducting sphere of radius $R$, containing a point charge $q$ a distance $a$ from its center. Thus we need to solve Poisson's equation with boundary conditions: $V(R)=0$ $V \rightarrow 0$ as $r \rightarrow \infty$ We consider an entirely different problem, which will hopefully solve ...


1

Write down an equation for the potential at any point in space. Set this equation equal to some constant C, and analyze the resulting relationship between x, y, and z. Compare this to the well-known equation of a sphere at the origin. EDIT: $$4\pi\epsilon_0V(x,y,z)=\frac{q_1}{\sqrt{x^2+y^2+(z-z_1)^2}}-\frac{q_1\sqrt{z_2/z_1}}{\sqrt{x^2+y^2+(z-z_2)^2}}$$ On ...


1

You're only solving for the field intensity above the plate. The "effects" of $-q$ there will be felt inside the region of $\epsilon$. From the perspective of that region of interest, $-q$ might as well be sitting in a region of $\epsilon$ aswell, the form of the magnetic field above the plate will not change.


1

why do I need to demand that the electric potential is continuous at the boundary, instead of just gauge it to zero inside the sphere. You need to demand that the electric potential be continuous at the boundary, because the physics requires it. You can, if you want to, demand that the electric potential be zero at the surface, if you find it to be ...


1

When I looked at it, I saw that when both charges are present, the force on a test charge at r is smaller in magnitude (because the test charge experiences two forces in opposite directions) and so the potential (which is integral F.dr) would be smaller than if only q1 were present. You are confusing two different quantities, force and potential. They are ...


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