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16

First remember that $k = \dfrac {1}{4 \pi \epsilon_r\epsilon_o}$ where $\epsilon_r \ge 1$ It is because a medium can be polarised by an external E-field. The dipoles so set up produce the external E-field produce an E-field in the opposite direction so the net E-field (the sum of the external and dipole produced E-fields) is smaller. Thus the force a given ...


10

In electromagnetism there are both positive and negative charges. Hence the force due to electric charges can be attractive or repulsive. Gravity, when treated as a classical force field, can only be attractive, there are not two types of "gravitational charge". What this means is that in electromagnetism, a given medium, may contain both positive and ...


5

Both gravity and electrostatic forces depend on distance ($r$) like $1/r^2$. So changing the separation between 2 atoms changes both forces equally. So whichever force is stronger initially (at any distance) will always be stronger. To determine which is stronger consider the ratio of gravitational to electric force. $$ F_g/F_e = 4\pi \epsilon_0 G ...


4

Just a coincidence. There are too many quantities and not enough letters. It probably does make a difference that the fields in which these two equations exist (material science and electromagnetism) are well enough separated that you typically won't see them both in the same papers or textbooks; if that weren't the case, people would start using different ...


4

In classical electrodynamics, assuming a point charge to be having a finite charge, the net electrostatic self energy carried by it is given by $$ Self Energy = 1/2 \int E^2 dV$$ Upon performing the intergral in three dimensions, since the electric field of a point charge diverges at the origin, therefore the rest mass by the virtue of the electrostatic ...


3

Electric field extends to infinity in the sense that no limit after which the field would vanish was ever found. It is natural assumption that simplifies things. Coulomb's law is consistent with this assumption, but there is no model that would explain Coulomb's law from anything simpler.


3

They accumulate in the corners. Free excess charges (electrons) in a conductor (metals) will move freely towards the surface until the internal electric field is zero. If this was not the case, the non-balanced field will further move them until this is the case. But the geometry of the dice makes the faces very unstable place for charges to accumulate, ...


3

Your intuition is fine - the particle will move just as you said. There are two problems with your reasoning, which lead to the conflict with your intuition. First problem: It seems that you're imagining that "zero potential energy" means that an object won't feel any force, and won't move. That simply isn't the case - it's the spatial derivative of the ...


3

You can't calculate it "directly" because it is not given. Rather it is the boundary condition of $V$ which is usually given. So you don't know what the induced charge distribution is in advance. If you don't want to use the image charge method, you have to solve the Poisson's equation with other methods first. Then calculate the E field, and then obtain the ...


3

I don't know the mathematical rigorousness to answer well the question but here's a thought. The left hand side is not 0. Take a point charge for simplicity, in this case $\vec E$ and $d \vec a$ are orthogonal and so the integral is worth $E\cdot A$. Although it is true that $\vec E$ tends to $\vec 0$ for $r \to \infty$, $A$ tends to infinity. And so you ...


3

No, atoms have the same number of protons and electrons so they have no net charge. On the other hand ions (cations and anions) would be repelled or attracted depending on their net charge. Atoms are bound together in a molecules by different means like covalent bonding, ionic bonding (which can be easily explained in terms of electrostatic forces) or ...


2

Assuming incandescent bulbs rather than, say, LEDs: they won't take "almost all" the energy, they'll take all of it. It doesn't matter if you use one bulb or five, it'll use up all of the energy that the battery is supplying. That's why multiple bulbs will be dimmer than one single bulb. Electrons collectively don't have a very good sense of how much energy ...


2

The motion of the charges is totally unlike you setting off from home walking 10 km and feeling a little tired and slowing down and then walking another 10 km and feeling even more tired and slowing down and then walking a final 10 km and arriving home at a very slow pace, indeed just reaching your front door and stopping. Here is a simple model of what ...


2

In your last question it is important as to what you mean by WRT the question. If you are trying to find the E-field due to a point charge using Gauss then to make the surface integration easier you choose a surface which has the following properties: the E-field direction is everywhere perpendicular to the surface the E-field has a constant ...


2

Consider a more general case in which the three charges are $\lambda q, q$, and $q$, with $\lambda\geq 0$. Let $A$, $B$, and $C$ be the respective positions of the charges. By symmetry, we have $|AB|=|AC|$. Due to the conservation of momentum, the center of mass $O$ of the system is stationary. Let $\theta$ be the angle $OBC$, and $\alpha$ be the angle ...


2

Take a simple 1D example. We'll plot gravitational potential energy as a function of horizontal position for the side of a hill: We normally define $F = -\nabla V$ and that means the force points downhill i.e. if we let the particle go it will roll downhill. If instead we defined $F = \nabla V$ the force would point uphill i.e. if we let the particle go ...


2

The tangent line, in general, is the line that follows the gradient of a curve at a specific point - like in this diagram: The blue line is the tangent of the red line at the point where they touch.


2

The field inside a conducting sphere is zero, not linear in $r$. The field of a uniformly charged sphere goes linearly with $r$.


2

As the charge is outside the surface the flux entering the surface is equal to the flux leaving the surface and total flux is therefore zero. The electric field is not zero. Flux =$\int \mathbf{E.ds}$ is zero not the electric field itself. for a negative single charge electric filed is $\mathbf{E}=\frac{Q}{4\pi \epsilon_0 r^2}(-\hat{r})$. The flux is zero ...


2

"Why?" questions are always a bit dubious in physics, but let me explain a few words here: "Voltage" is in itself a difference, and it is (by definition) the difference of electric potential energy. The difference in electric potential energy is by definition the work needed to move a test charge against the electric field. In other words: The electric ...


2

Here is how: And the expression is (from wikipedia): $$2\pi \varepsilon a\sum_{n=1}^{\infty }\frac{\sinh \left( \ln \left( D+\sqrt{D^2-1}\right) \right) }{\sinh \left( n\ln \left( D+\sqrt{ D^2-1}\right) \right) } $$


2

Yeah, that's just a coincidence. The easy way to see this is that $\epsilon$ is a relatively static property of a dielectric but a totally dynamic property of a stretching material.


2

Engineers created that problem. ;) (probably not) Many physics books use $Y$ for Young's modulus (Symon, Knight, Young & Freedman). Taylor's Classical Mechanics uses YM. Halliday, Resnick & -fill-in-the-blank- state that engineers use $E$. I suspect that physicists started using $Y$ for exactly this reason: to highlight a difference in the meanings ...


2

q1 and q2 induce charges of the same magnitude and of the opposite sign on the surface of the spherical cavities. This in turn means that on the outside surface of the sphere there are charges induced of the same magnitude and the same sign as charges q1 and q2. These charges are distributed on the outside surface of the spherical conductor. That is there ...


2

There many configurations that statisfy your assumptions. But you had forgot about many other constraints like potential difference, charge density, end effects etc. When you consider PD you come to know, ignoring end effects, that only one configuration is possible. Since 1st plate has charge Q its surface charge density of 1st side is Q/2A 2nd side again ...


2

You're only a half-step away. You listed conservation of energy and linear momentum, both of which are due to there being no external forces on the three-charge system. But with no external forces, you know that the center of mass of the system won't accelerate. Since the COM starts at rest, this means that the COM will remain stationary. Think about what ...


2

It might help you to think about the symmetry of the situation. First in the application of conservation of momentum and then what the trajectories of the charges must be to keep the centre of mass $C$ at the same position. This will give you a connection between $v_y$ and $v_x$.


2

Actually we have models explaining this. The particles that mediates electromagnetic field are massless, so the range of the force predicted by the model is infinite. On the contrary, for massive mediators (see for instance Yukawa force), the range is finite. Notice that real experimental setups have finite precision, so beyond some limit it's pointless to ...


1

In a copper wire there are copper ions with 28 bound electrons orbiting the nucleus. Approximately one electron per copper atom is free to roam throughout the metal and these electrons are called unbound or free electrons ans are responsible for electrical (and heat) conduction in copper. The positive copper ions are bound together and only vibrate about a ...


1

The direction is correct, just note that you can write $$ \frac{\partial^2\phi}{\partial x^2} = - \beta^2\phi $$ and $$ \frac{\partial^2\phi}{\partial y^2} = \beta^2\cdot\left( \phi -\frac{\rho_0}{\epsilon_0\beta^2}\cos{\beta x}\right) $$ This can ease the calculations. The result you obtain is an oscillating potential in $x$, not depending on $y$. The ...



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