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While the formula $V=E\cdot d$ works for parallel plate capacitor, it does not apply for an infinitely long rod. The Formula for calculating potential is $V=\int \vec{E}\cdot\mathrm{d}\vec{l}$. And be aware that we can no longer choose infinity as the zero potential point because the rod itself is infinitely long.


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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 ...


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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 ...


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Appending to the answer given above, since I cannot post comments. Integrating in the onion peel manner, charge enclosed in the gaussian surface of a 'thin' shell taking care of limits (inner radius and beyond to whatever distance so desired) should yield you the answer to your question. Note the exponent of the distance parameter in your 'distance - ...


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To answer the first question, consider that if the electric field acts in one dimension, then $\mathbf{E} = -\frac{d\mathbf{V}}{dx}$ Hence, in a uniform field the potential changes linearly across the direction parallel to the field. Like this image, where the potential increases by a constant amount at each equipotential as we go left (because field ...


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One simple explanation is that the work you calculate does not depend on the path along which you move the charge. Or, in other words, if you move a charge in a closed path, the work done by the electric field is zero. See, $A=\oint q\vec E\cdot d\vec r=qQ(1/r - 1/r)=0$


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You only have an electric field if the current that passes is not neutral(i.e. it has more positive or negative charges in a unit length) or if the current changes with time. In the second situation,the magnetic field around the changes with time(so is the magnetic flux inside a "made-up" area around the wire),so nature causes an induced electric field in ...


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"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 ...


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we know, $$E=V/d$$ (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html ) Where $V$,$E$ and $d$ are potential difference ,electric field and distance between the two plates respectively. Electric field between the plates only depends on charges of the plates and since charges must be conserved so when the plates are moved apart charges(the ...


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I believe that the photons of EMR "escape" the source of them (excited electrons), whereas somehow the photons in a magnetic field just circle around.


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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.


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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 $\frac{1}{L}$


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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} ...


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Then there would be electric field inside that shell. Well if it is still intact. Because you need an immense positive charge to do that to a typical shell, more than 100000 Coulombs (assuming you have 1 mole of conductor with about 50 electrons per atom). To bring that much charge together requires immense force and energy. In fact potential energy of 10 cm ...


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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 ...


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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 ...


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 ...


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A test charge is an idealization - an infinitesimally small charge. A real charge causes redistribution of charge on the metal ball, so more negative charge moves closer to the external positive charge, so the measured force becomes greater than it would be if the charge were not redistributed, and it is the non-redistributed charge that determines the ...


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The general equation for electric field due to dipole at a point is KP/R^3[(1+3cos^2(thetha)]^1/2. When (thetha=90 it is equitorial line and when (theta=0) it is axial line from the above equation we get the required results ie KP/R^3 and 2KP/R^3


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We split $\rho=\rho_{\it ex}+\tilde\rho$ and $\jmath=\jmath_{\it ex}+\tilde\jmath$ into "explicit" and "medium" contributions. We can then define $P$ and $M$ by $\tilde\rho=-\nabla P$ and $\tilde\jmath=\nabla\times M$, and macroscopic fields by $D=E+P$ (I'm sloppy about units, there is a $4\pi$ in Gauss units) and $B=H+M$. The macroscopic equations are then ...


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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 ...


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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 ...


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as you know that inside a capacitor electric field remains same. If you increase the distance between the two plates electric field does not change just because electric field= surface charge density/ epsilon. so E=V/D gives increment in V as D increses so that electric field remain same.


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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

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 ...


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Work done by the electric field is force times displacement along the direction of force. Center of mass doesn't move, so the only motion is rotation; horizontal spacing of charges goes from $a\cos\theta$ to $a$, so work done is $E q a(1-\cos\theta)$. Half of that work is done on the positively charged mass (which moves half that distance, i.e. ...


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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 ...



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