The divergence operator and Gauss’s’law
The divergence operator and Gauss’s’law

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Divergence and Curl of Electric Field
Consider an electric field E = c – – (Note the numerator is not âˆš: this is not the usual E field from a point charge at the origin, which would give c).
a. Calculate the divergence and the curl of this E field. Explicitly test your answer for the divergence by using the divergence theorem. (Is there a delta function at the origin like there was for a point charge field, or not?)
b. What are the units of the constant c? What charge distribution would you need to produce this electric field? Describe it in words as well as formulas. (Is it physically possible?)

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06:18

Gauss’ Law for electric fields The electric field due to a point charge $Q$ is $\mathbf{E}=\frac{Q}{4 \pi \varepsilon_{0}} \frac{\mathbf{r}}{|\mathbf{r}|^{3}},$ where $\mathbf{r}=\langle x, y, z\rangle,$ and $\varepsilon_{0}$ is aconstanta. Show that the flux of the field across a sphere of radius $a$ centered at the origin is $\iint_{S} \mathbf{E} \cdot \mathbf{n} d S=\frac{Q}{\varepsilon_{0}}$b. Let $S$ be the boundary of the region between two spheres centered at the origin of radius $a$ and $b$ with $a<b$, Use the Divergence Theorem to show that the net outward flux across $S$ is zero.c. Suppose there is a distribution of charge within a region $D$. Let $q(x, y, z)$ be the charge density (charge per unit volume). Interpret the statement that$\iint_{S} \mathbf{E} \cdot \mathbf{n} d S=\frac{1}{\varepsilon_{0}} \iiint_{D} q(x, y, z) d V$d. Assuming E satisfies the conditions of the Divergence Theorem, conclude from part (c) that \$\nabla \cdot \mathbf{E}=\frac{q}{\varepsilon_{…

00:33

Consider a spherical Gaussian surface of radius R centered at the origin: charge Q is placed inside the sphere. To maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located:A) at the originB) at x=0, y=0, z=R^2C) at x=0, y=R^2, z=0D) at x=R, y=0, z=0E) The charge can be located anywhere, since flux does not depend on the position of the charge as long as it is inside the sphere.

02:09

A nonuniform but spherically symmetric distribution of charge has charge density p(r) given as follows:For r < R: p(r) = poFor r â‰¥ R: p(r) = plrwhere po = 3aL Tk and R is the radius of the charged sphere.

What is the electric field as a function of radii greater than R (in terms of Qtotal and the distance from the center of the distribution)?

By integrating the charge over the entire region of the sphere, show that po = 3Qpl.

Find the electric field as a function of the inside of the sphere using Gauss’ Law.

01:40

Consider a point charge q in three-dimensional space.Symmetry requires the electric field to point directly away fromthe charge in all directions. To find E(r), the magnitude ofthe field at distance r from the charge, the logicalGaussian surface is a sphere centered at the charge. The electricfield is normal to this surface, so the dot product of the electricfield and an infinitesimal surface element involves cos(Î¸)=1.The flux integral is therefore reduced to âˆ«E(r)dA=E(r)A(r),where E(r) is the magnitude of the electric field on theGaussian surface, and A(r) is the area of thesurface.(Figure 1)

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