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5

...why isn't the work done against the net force due to the system considered instead of simply adding up the work done against separate forces caused by individual charges? They're both equivalent, due to the principle of superposition. Basically, the net force is what you get when you add up the separate forces from the individual charges acting on ...


4

You evidently understand that any constant can be added to a potential without affecting the physics -- or equivalently, any place can be taken to have zero potential. You also suggest, rightly, that there are really only two "natural" places to define the zero of the potential: either $r=\infty$ or $r=0$. For example, there's no particular reason to ...


3

You have a sign error. The potential is defined by see: http://en.wikipedia.org/wiki/Electric_potential


2

Basically, the system is simply that which is studied in a problem in physics. It refers to that which we want to know more about, in this case the moving electric charge in the presence of the electric field. Be cautious with the terms 'electric potential' and 'potential energy', since they're two different things. Electric potential is defined as ...


2

Observe the potential lines for a moment. You will find that for equal change in distance, there is equal change in potential. Means, if I move 0.5 m to the left, the potential increase is 10 V. In other words, we have equidistant equipotential lines which is a graphical way of denoting uniform field. Whenever you see straight equipotential lines, it means ...


1

I'm going to have to give an answer that's very different to Jimmy360's. Apologies. How does the potential and kinetic energy of a photon relate? They don't. The photon is all kinetic energy. Do they mean the same thing? No. When you drop a brick, its gravitational potential energy is converted into kinetic energy. When you dissipate this kinetic ...


1

Because they should have the same electric potential and electric potentials of them depends on charge not charge density.


1

The potential at the surface of a charged sphere or radius $r$ is: $$ V = k\frac{Q}{r} \tag{1} $$ Since the area of the sphere is $4\pi r^2$, the charge density is: $$ \rho = \frac{Q}{4\pi r^2} $$ and rearranging gives: $$ Q = \rho 4\pi r^2 $$ and substituting for $Q$ in equation (1) we get:$$ V = 4\pi k\rho r $$ Can you take it from here?


1

On the one hand, you can't solve for the magnetic field without appropriate boundary conditions (e.g. there could always be an incoming electromagnetic wave that hasn't yet impinged on your cylinder). On the other hand if you have a fixed charge and current distribution you can always use Jefimenko's equations to find a solution to Maxwell's equations, and ...


1

For a classical point charge, the field is divergent at $r=0$, and if you were to take the potential to be zero there, it would be infinite everywhere else. Meanwhile, you can approximate $r=\infty$ as the region with no interaction, so it's reasonably naturally to treat it in the way you would treat ground.


1

You are right that the charge gained potential energy. But this statement is only true because the charge is part of a system. We cannot talk about the electrical potential of a charge unless it is in an electric field - which means that there is "something else" that is essential for our definition of the potential. We say the "system" (the charge, plus ...


1

The key here is that E must exceed the minimum of $V(x)$. If one has a delta function well then the minimum is $V_0 = -\infty$ while for a delta function barrier the minimum is $V_0 = 0$ Hence for the delta function well, one can have bound states with $E<0$ or one can have scattering states with $E>0$. For the delta function barrier the necessary ...


1

This is accurate, and it ultimately comes down to the fact that we can get arbitrarily close to an electric point charge in classical E&M. That means that the field right up next to the point charge could be arbitrarily large. So you get these huge, singular potentials close to point charges, which is really more-or-less fine. For instance, that huge ...


1

You have made two elementary errors. First, you forgot a negative sign and second, you forgot your limits (V at infinity is conventionally taken to be zero) $V = -\int_{\infty}^{r} E dr$ Since I believe you are taking the electric field at the x axis, instead of r you can use x, making that part of your work correct. These two things will fix your ...


1

Since the charge Q have to be the same for both capacitors and you need more voltage to to push that charge in the capacitor with less capacitance then you must have more valtage difference in $C_2$ The mechanical analogy is a configuration with 2 springs in parallel that move the same distance from their equilibrium position need more force on the spring ...


1

The working voltage of a capacitor depends on the dielectric strength of the insulator. While electrical breakdown is actually a very complicated process with lots of non-linearities, you can simplify the design of a capacitor by saying "the electric field on the insulator must not exceed X". Once you have said that, and you realize that the electric ...


1

I think your problem is that you're adding the electric fields like scalars, rather than breaking then into their vector components. You have: $ E_1 = \frac{q}{2\pi\varepsilon_0\rho_1} $ and $ E_2 = \frac{-q}{2\pi\varepsilon_0\rho_2} $ What you should have is $ E_1 = \frac{q}{2\pi\varepsilon_0}(\frac{1}{\rho_{1x}}\hat{x} + \frac{1}{\rho_{1y}}\hat{y})$ and ...


1

Since this is a homework problem, I won't provide a full solution, but here's a nudge in the right direction. Take a look at these two plots of the effective potential: k = -1, $\alpha$ = 1, L = 0.25 k = -1, $\alpha$ = 1, L = 1 What's different about these two effective potentials? We only changed $L$ between the two graphs; what does that imply about ...


1

I think the electric field is zero on A, B, C, D and E, because otherwise there would be current, which would be odd And you are totally right for an electrostatic system (with no current). Instead of explaining it by, "this would be odd", let's have a look at what happens in the instant you add the wire to the battery pole.: Before the wire touches ...


1

For a diode, no voltage means no current (and vice versa). That is, an operating point of a diode is (0 V, 0 A). So with $I_c=0$, there is no voltage dropped across the diode, so point $A$ is at the same potential as $V_{cc}$


1

I'm assuming you mean a solid metallic sphere, not a shell. The analysis is slightly different for a shell but rests on the same principles. It is actually not correct that the total charge induced on the sphere in the case of grounding is $-q$. The basic concepts you need to know about electrostatics with conductors is that the entirety of the conductor ...


1

I will presume that, as is standard and conventional, the electric potential at a place an infinite distance from the one in consideration is zero. Let the charge at point a be charge $a$, the charge at point b be charge $b$ and the charge at point c be charge $c$. With this, the formula for the electric potential is: $V = ...


1

What the voltmeter is reading in the top circuit would be, if you used a Kirchhoff's loop rule in the loop containing 2 resistors and the voltmeter, the difference in the potential difference across the top-right resistor and the bottom-right resistor. The loop rule calculation would look something like this ($A,B,C,\&D$ are the top-left, top-right, ...



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