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6

Electrostatic refers to the case where the fields are not time dependent. In that case the Maxwell's equations reduce to: $$\nabla \cdot E =\frac{\rho}{\epsilon_o} \\ \nabla \times E = 0 \implies E=-\nabla \phi \\ \text{then,} \nabla \cdot \nabla \phi = \nabla^2 \phi = -\frac{\rho}{\epsilon_o} $$ The solution to the last equation is: $$ \phi = ...


5

The reason is the samen as why the electric field inside a conductor is zero: if it isn't zero, the free electrons undergo a force and move (rearrange) untill they dont feel a force anymore. If the electrons don't feel a force, the electric field must be zero. At the surface of a conductor, the free electrons feel a force perpendicular to the surface, but ...


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 ...


2

Depending on the location of the switch, the answer will change. A properly wired lamp would have no signal on the live (phase) wire, and therefore there would be no field. However, if you interrupt the neutral wire (or the switch is in the lamp, not the wall) then you will have a varying AC field because the voltage on the wire changes (and thus a small ...


2

Your first two equations are effectively the same once you consider that $F = q E$. You should use the second one if you're concerned about the electric field in general, but the first one if you want to know what actually happens to a particle (i.e., what force a particle feels). When in doubt, you can usually do alright by working out first the field, and ...


2

You say the lamp is plugged into a AC outlet, but then talk of a "wall switch". Apparently you mean that this switch controls the power to the outlet, and that a switch on the lamp is kept on, or that the lamp has no switch. If so, you should clarify this as a switched AC outlet, since most aren't. In the case of a switched AC outlet, the switch will be ...


2

Prove I have taken that line charge is placed Vertically and one test charge is placed. Now the electric field experienced by test charge dude to finite line positive charge. $$Ex = \int dx cos \alpha$$ $Ey$ will be cancel out as they will be opposite to each other. $$Ex = \int k \frac{dq}{x^2+y^2}cos\alpha$$ $$Ex = \int k \frac{\lambda ...


1

The field strength is only $F/Q$ if the test charge is infinitesimally small so it cannot affect the metal ball. For example, suppose the metal ball is uncharged in which case the fortce should be zero. A positive charge will polarise the ball and create an attractive force even though the charge on the ball is zero. In the case of a charged ball the ...


1

If there were to exist an electric field in the conductor, there would also have to be a current. A radial field drives a radial current, and it is this very current that will change the charge that is on either surface, which then in turn creates its own electric field. The steady-state is when the TOTAL field is equal to zero, when there are no longer any ...


1

The point is: what do you call a "conductor"? If you are talking about a medium which follows Ohm's law $ \vec{j} = \sigma \vec{E} $, then $\vec{E}$ has no reason to equal zero (well, in steady state, $\vec{j}=\vec{0}$ so...). But if you are dealing with a perfect conductor, in which $\sigma \rightarrow +\infty $, then $\vec{E} = \vec{0}$: $$ \begin{cases} ...


1

You have to change Q in the integral and substitute it for the differential charge on one of those infinitesimally small rings. I think that should change your result only by a factor of $\frac{1}{L}$


1

Two electric fields line can never cross each other because at every point there is unique tangential direction of electric Fields.if they did, field at the point of intersection will not have unique direction, which is absurd.


1

"So if I construct a gaussian surface such that it extends out from both the sides ,then I should get the same answer as that for a charged sheet! Why do we not do this in this case ?" if you do that you have to realize that the other side have same amount of charge as the first, so the filed is doubled As for your last question if you consider a ...


1

The basic concept is to see the charge distribution among them. From these example I hope your doubt will be quite clear First left us take one non-conducting sheet then, $$E ds + E ds + 0 = \frac{1}{E_o}[\sigma.ds] $$ For curved surface it will be zero and let us take sigma distribution on sheet. See Figure 1. Now for conducting sheet (Case 1) Here ...


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

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

The electrons are in random motion within the cord even when it is plugged and not switched on. The motion of the electrons in this is case is random i.e., there is no preferred direction of motion of electrons or vector sum of all the thermal velocities is zero. Each electron within this conductor acts like a point source of electric filed and these micro ...


1

Since you know that the field inside a conductor is zero, you can apply Gauss' Law for flux to say that any spherical surface lying inside the conductor cannot have any flux through it, so that the enclosed charge is zero. Therefore, you need -2.3mC of charge from the conductor on the inner surface to make sure that the enclosed charge is zero. For the ...


1

The surface of the conductor has constant electrostatic potential $V$ and the electric field is proportional to the gradient of the potential: $\nabla V$. By definition the gradient of a scalar quantity is always perpendicular to the level curves (surfaces).


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Pay close attention to how you defined $x$: it's not the coordinate of the desired point on the $x$ axis, it's how far left of the positive charge that point is. So when your equation tells you $x = 4.83$, that means the desired point is $4.83$ units left of the positive charge. But, presumably the question you've been asked wants the coordinate of the ...


1

In my opinion, your solution is correct. The book's solution is correct as well, but you would need to draw a different diagram for it to make sense. Their assumption is to take by default, r as lying on the positive x axis. In this way, the distance between the $Q_2$ and the required point is the mod of $r-2$. Since this expression is being squared, it ...


1

You can find the expression for the electric field of a finite line element at http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html - which gives for the Z component of the field of a line that extends from z=a to z=b $$E_z = \frac{k\lambda}{z}\left[\frac{b}{\sqrt{b^2+z^2}} + \frac{a}{\sqrt{a^2+z^2}}\right]$$ You can follow the approach in that ...



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