Suppose the electric field in some region is fo...

Suppose the electric field in some region is found to be $E=kr^3\hat{r}$ , in spherical coordinates (k is some constant).

(a) Find the charge density ρ.

(b) Find the total charge contained in a sphere of radius R, centered at the origin. (Do it two different ways.)

Question Data is a breakdown of the data given in the question above.

(a) The electric field in spherical coordinates is given by E = kr^3.

(b) The problem asks to find the charge density ρ.

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Step 1:

To find the charge density ρ, we can use the equation ρ = ε0 ∇⋅E, where ε0 is the permittivity of free space.

Step 2:

Using spherical coordinates, we substitute the given electric field equation E = kr^3 into the charge density equation.

Step 3:

We calculate the divergence of the electric field E using the spherical coordinate system.

Step 4:

Taking the derivative of the radial component r^2⋅kr^3 with respect to r, we simplify the equation.

Step 5:

Simplifying further, we find that the charge density ρ is equal to 5ε0kr^2.

Step 6:

To find the total charge contained in a sphere of radius R, centered at the origin, we can use two different methods: Gauss's law and direct integration.

Step 7:

Method 1 - Gauss's law: We apply Gauss's law, which states that the total electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space.

Step 8:

Using Gauss's law, we calculate the electric flux (∮E⋅da) through a sphere of radius R, centered at the origin.

Step 9:

Substituting the given electric field E = kr^3 into the equation, we simplify the expression.

Step 10:

Further simplification yields the total charge enclosed (Qenc) as 4πε0kR^5.

Step 11:

Method 2 - Direct integration: We find the total charge by directly integrating the charge density ρ over the volume of the sphere.

Step 12:

Setting up the integral, we multiply the charge density ρ by the differential volume element (r^2⋅dr⋅dθ⋅dϕ) and integrate over the appropriate limits.

Step 13:

Evaluating the integral, we simplify the expression to obtain the total charge enclosed (Qenc) as 4πε0kR^5.

Note: This explanation provides a step-by-step approach to solving the problem, explaining the reasoning and formulas used at each step. It is important to follow each step carefully to ensure an accurate solution.

Final Answer

(a) $\rho =\varepsilon _0∇\cdot E=\varepsilon _0\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 · kr^3\right)=\varepsilon _0\frac{1}{r^2} k \left(5r^4\right) =5\varepsilon _0kr^2$ .

(b) By Gauss’s law: $Q_{enc}=\varepsilon _0\oint{} E·da =\varepsilon _0\left(kR^3\right) \left(4\pi R^2\right)=4\pi \varepsilon _0kR^5$ .

By direct integration: $Q_{enc}= \int{}\rho d\tau =\int_{0}^{R}{}\left(5\varepsilon _0kr^2\right)\left(4\pi r^2dr\right)=20\pi \varepsilon _0k\int_{0}^{R}{} r^4dr=4\pi \varepsilon _0kR^5$ .