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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 ...


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Does the electric field at a point on the surface change? Yes, it change. Where distance is decreased Electric field increase. Does the total flux through the Gaussian surface change? No, it remains same. Consider a spherical gaussian surface of radius $2m$ and a charge $q$ inside it. Let this charge emit $10$ field lines. Then, clearly, all the field ...


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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 ...


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Your approach using Gauss and the integration to find the total charge is correct. The method which just multiplies the charge density as though it was constant (independent of R) by the volume of a sphere is incorrect. The E-field lines will be radial and be in the outward direction if the charge is positive.


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Well you can find approximate values by reasoning but for the exact value you'll need at least to know limits. Let's reason for a moment as you are proposing, but slightly different. Since the plane is infinite you can consider the point in the center of the plain, and at distance R from it. Now think of the plain as a collection of rings of very fine ...


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How about this? 1) There is a charged spherical shell. The origin of the sphere must not have any electric field due to symmetry. $${\bf E}({\bf 0}) = {\bf 0}$$. 2) Now take a point from the to the origin at $\bf r$. Due to symmetry of the problem the electric field has to be radial (points away from the origin), but can (still) have a magnitude $A_r$. ...


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So, this is an interesting property of the mathematics of a force that diminishes like $1/r^2$ in 3D-space: if you have a uniform charge distributed over a sphere, that charge exerts no forces inside the sphere; they all balance out. Furthermore the field outside the sphere behaves as if all of the charge on the sphere was concentrated at a point at the ...


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Since there are no charges inside a charged spherical shell . This means the net charge is equal to zero. So magnitude of electric field $E$=0. So there is no net force. So magnitude of net force =0. No movement towards centre.


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It would experience no net force inside shell. It is free to move.


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how is the divergence of electric field related to the electric flux? You are looking for Gauss' Law, which should be in whatever textbook you are using as a guide to this subject: $$\int E\cdot dA=\frac{q}{\epsilon_0},\\\frac{\rho}{\epsilon_0}=\nabla\cdot E$$


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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 ...



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