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The flat ends of a cylinder are perpendicular to its cylindrical surface. The electrical field is perpendicular to any cylindrical surface centered on the line charge, and so is parallel to the ends of any such cylinder. $\vec E$ is the electric field, and I presume $\vec n$ is the surface normal of the ends of the Gaussian surface. Since the two vectors ...

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This is one of the places where we can make things perfectly rigorous if we make certain assumptions on the charge density $\rho$ (and $D$). I will rigorously show you in the following that $\Vert \mathbf{E}(\mathbf{x}_{0}) \Vert < \infty$ for all $\mathbf{x}_{0}$ in the interior of $D$, assuming that $\rho \in (L^{1} \cap L^{\infty})(D)$. The only ...

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You have more flux per unit area going into the right side, but the area on the right side is smaller. These two balance out so that the total flux is the same going in as going out. The part of the sphere which has electric flux going in, traced in red, is less than half the area of the sphere. Incidentally, flux per unit area is just the electric ...

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It is true that it is much stronger on the side the flux is into the sphere, however if you imagine drawing a cone whose tip is at the charged point and such that the cone is tangent to the sphere (imagine an ice cream ball in a cone) then you will realize that the closer the charge is to the sphere and the stronger it there fore is on the near side, the ...

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It appears to me that you are slightly confused with regards to the concept of current in conductors. Now, if I only choose one side of this rectangle, and apply external electrical field ∑ only to it, what EMF would I create on the conductor? I would simply say ∑, however then I had the following idea, and I started to doubt if I create 2∑ instead ...

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You know that if you have a point charge with charge $Q$, then the potential difference $V$ between spatial infinity and any point a distance $r$ from the charge is given by $$V_\textrm{point}=\frac{kQ}{r}.$$ You also know that the electric field from an infinite sheet of charge with charge density $\sigma$ is given by $$E_\textrm{sheet}=2 \pi k \sigma.$$ ...

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It is true that given an electric field, then you can define uniquely the charge density that created it, by Gauss' law, as you have done. But the converse is not true: given a charge density you cannot define uniquely the electric field that it will create since you have to solve a differential equation (again Gauss' law) to do that and each differential ...

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calculate centre of charge by the process of integration and then find potential and fields.

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The integral of $sin(\theta)$ from 0 to $2\pi$ is zero. We can calculate it by $\int_{0}^{2\pi} sin(\theta)d\theta = [-cos(\theta)]\Big|^{2\pi}_{0} = -cos(2\pi) + cos(0) = -1 + 1 = 0$. So when you integrate to find $E_{r}$, you will find that it equals zero at the center of the circle. Also, you mentioned the integral from 0 to $\pi$ is 2. ...

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An alternative and shorter answer is that the expression you cite for $\bf{E}(\bf{x}_0)$ makes use of the Green function $\bf{G}(\bf{x}; \bf{y}) = \bf{G}(\bf{x} - \bf{y})$ satisfying the distributional equation $$\nabla_{\bf{x}} \cdot \bf{G}(\bf{x}-\bf{y}) = \delta(\bf{x} - \bf{y})$$ and reading $${\bf G}(\bf{x} - \bf{y}) = \frac{\bf{x} - ... 1 Let a square conductor frame length l, resistance R,be pulled out with a constant velocity v from a magnetic field B perpendicular to plane of the frame .Then an emf e=Blv is produced across the frame. The EMF is around a closed loop. It is equal to the force per unit charge around the loop. So for instance$$\mathscr E=\oint_{\partial S}\left(\vec ...

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I believe the following picture explains what's missing: The cylinder is "infinite", but the Gaussian surface that is drawn as part of the analysis is in the shape of a finite cylinder with flat ends. And since the electric field is at every point perpendicular to the wire, it is parallel to these flat ends. Parallel to the surface means perpendicular to ...

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One thing that might help is to be clear about what is a vector and what is a scalar. You've done your integration with all the $y$ components of your vector. All you need to do is perform a similar integration for both $x$ and $y$ components.

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