Question

A non-conducting spherical shell, with an inner radius of R1 and an outer radius of R2, has carge spread non-uniformly through its volume between its inner and outer surfaces, where the volume charge density r is given by r=k/r with the distance r frome the center of the shell and a constant k.
(a) Find the net charge Q of the shell
(b) Calculate the electric field E at a distance R frome the center over the range 0 ￡R￡￥
(c) Fin the maximum E.

Answer 1

Answer on Question # 70191, Physics / Electromagnetism

Question. A non-conducting spherical shell, with an inner radius of $R_{1}$ and an outer radius of $R_{2}$ , has charge spread non-uniformly through its volume between its inner and outer surfaces, where the volume charge density $\rho$ is given by $\rho = k / r$ with the distance $r$ from the center of the shell and a constant $k$ .

(a) Find the net charge $Q$ of the shell;

(b) Calculate the electric field $E$ at a distance $R$ from the center over the range $R_{1} < R < R_{2}$ ;

(c) Find the maximum $E_{max}$ .

Given.

$R_{1}$

$R_{2}$

$\rho = k / r$

Find.

Q;

$E_{r}$

$E_{max}$

Solution.

a) Let us consider a sphere of radius $r > R_1$ (see figure) then charge, enclosed by the considered sphere,

$$Q = \int_ {R _ {1}} ^ {R _ {2}} 4 \pi r ^ {2} \cdot \rho \cdot d r = \int_ {R _ {1}} ^ {R _ {2}} 4 \pi r ^ {2} \cdot \frac {k}{r} \cdot d r = \int_ {R _ {1}} ^ {R _ {2}} 4 \pi r \cdot k \cdot d r = 4 \pi k \cdot \frac {r ^ {2}}{2} \left| \begin{array}{l} R _ {2} \\ R _ {1} \end{array} \right. = 2 \pi k \left(R _ {2} ^ {2} - R _ {1} ^ {2}\right).$$

b) Now, applying Gauss' theorem,

$$E _ {r} \cdot 4 \pi r ^ {2} = \frac {Q _ {\text {e n c l o s e d}}}{\varepsilon_ {0}} \text {(w h e r e} E _ {r} \text {i s t h e p r o j e c t i o n o f e l e c t r i c f i e l d a l o n g t h e r a d i a l l i n e)}.$$

For $0 < r \leq R_{1}$ :

$$Q _ {e n c l o s e d} = 0 \rightarrow E _ {r} = 0.$$

For $R_{1} < r \leq R_{2}$ :

$$\begin{array}{l} Q_{enclosed} = \int_{R_1}^{r} 4\pi r^2 \cdot \frac{k}{r} \cdot dr = \int_{R_1}^{r} 4\pi r \cdot k \cdot dr = 4\pi k \cdot \left. \frac{r^2}{2} \right| \frac{r}{R_1} = 2\pi k \left(r^2 - R_1^2\right); \\ E_r = \frac{2\pi k \left(r^2 - R_1^2\right)}{4\pi r^2 \varepsilon_0} = \frac{k \left(r^2 - R_1^2\right)}{2 r^2 \varepsilon_0} = \frac{k}{2 \varepsilon_0} \left(1 - \frac{R_1^2}{r^2}\right) \end{array}$$

For $R_2 < r$:

$$Q_{enclosed} = 2\pi k \left(R_2^2 - R_1^2\right);$$ $$E_r = \frac{2\pi k \left(R_2^2 - R_1^2\right)}{4\pi r^2 \varepsilon_0} = \frac{k \left(R_2^2 - R_1^2\right)}{2 r^2 \varepsilon_0} = \frac{k \left(R_2^2 - R_1^2\right)}{2 \varepsilon_0} \cdot \frac{1}{r^2}.$$

c) As magnitude of electric field decreases with increasing $r$ for $R_2 < r$, field will be maximum for $R_1 < r \leq R_2$. Now, for $E_r$ to be maximum

$$E_{\max} = \frac{k}{2 \varepsilon_0} \left(1 - \frac{R_1^2}{R_2^2}\right).$$

Answer.

a) $Q = 2\pi k \left(R_2^2 - R_1^2\right)$.

b) for $0 < r \leq R_1$ $E_r = 0$;

$$\text{for } R_1 < r \leq R_2 \quad E_r = \frac{k}{2 \varepsilon_0} \left(1 - \frac{R_1^2}{r^2}\right);$$ $$\text{for } R_2 < r \quad E_r = \frac{k \left(R_2^2 - R_1^2\right)}{2 \varepsilon_0} \cdot \frac{1}{r^2}.$$

c) $E_{\max} = \frac{k}{2 \varepsilon_0} \left(1 - \frac{R_1^2}{R_2^2}\right)$.

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