Distributed (Volumetric) Charge Examples

A very long pipe has a distributed charge within it’s volume as $\rho (r)=D{\frac{r}{a}}^5$ where D is a constant, a is the inner, and b is the outer radii.

for $a<R<b$

$${\varphi }_{enc}=4\pi k{Q}_{enc}$$ $$Q_{enc}(r)={\int }_a^{R}2\pi rlD{\frac{r}{a}}^{5}dr$$ $$Q_{enc}(r)=\frac{2\pi lD}{a^5}{\int }_a^R{r}^{4}dr$$ $$Q_{enc}(r)=\frac{2\pi lD}{7a^5}(R^7-a^7)$$ $$E(r)=\frac{2k}{r_{1}l}\frac{2\pi lD}{7a^5}(R^7-a^7)$$ $$E(r)=\frac{2k}{r}\frac{2\pi vD}{7a^5}(R^7-a^7)$$

for $R>b$

$$E(r)=\frac{2k}{r}\frac{2\pi vD}{7a^5}(b^7-a^7)$$

Point charge in solid spherical shell

Find $E(r_1)$ for $r_1=2cm$

$$E(r_1)=\frac{k{Q}_{enc}}{r_1^2}$$ $$E(r_1)=\frac{8.9875\cdot 10^9\cdot 8\cdot 10^{-9}}{4}$$ $$E(r_1)=\frac{8.9875\cdot 8}{4}=17.975$$

Find $E(r_2)$ for $r_2=4cm$

$$E(r_2)=0\text{for}3cm<r_2<9cm$$

When charges are static, there can be no electric field inside a conductor.

Find $E(r_3)$ for $r_3=11cm$

$$E(r_3)=\frac{k{Q}_{enc}}{r_3^2}$$ $$E(r_3)=\frac{8.9875\cdot 10^9\cdot -5\cdot 10^{-9}}{11^2}$$ $$E{r}_3=-0.371384297521$$

Calculate the surface charge density on the inner and outer surface of the iron shell

Due to the central charge, $Q_{inner}=-8nC;Q_{outer}=8nC$

The additional $-13nC$, will drift to the outer surface, giving:

$$Q_{inner}=-8nC;Q_{outer}=-5nC$$ $${\Sigma }_{inner}=\frac{-8\cdot 10^{-9}}{4\pi 0.09};{\Sigma }_{outer}=\frac{-5\cdot 10^{-9}}{4\pi 0.81}$$

Varied charge over volume sphere

$$\rho =a{r}^2\text{for}r<R$$

What is the electric field when $r>R$?

$$E=\frac{q_{enc}}{4\pi r_a^2{ϵ}_0}$$ $$q_{enc}={\int }_0^{R}4\pi r^2\cdot a{r}^{2}dr$$ $$q_{enc}=4\pi a{|}_0^R\frac{r^5}{5}$$ $$E=\frac{4\pi a{|}_0^R\frac{r^5}{5}}{4\pi r_a^2{ϵ}_0}$$ $$E=\frac{a}{r_a^2{ϵ}_0}\cdot \frac{R^5}{5}$$

Rod in Pipe

$R_1=a/2$, $R_2=3c$ Find $V(R_1)-V(R_2)$

$$V(R_1)-V(R_2)=-{\int }_{R_2}^{R_1}Edr$$

Split integral into appropriate continuous regions:

$$=-{\int }_{R_2}^{c}Edr-{\int }_c^{b}Edr-{\int }_b^{a}Edr-{\int }_a^{R_1}Edr$$

Electric fields inside of surfaces are zero:

$$=-{\int }_{R_2}^{c}Edr-{\int }_b^{a}Edr$$

Apply Guass’s Law for $R_2\to c$:

$$Flux=4\pi rk{Q}_{enc}$$ $$EA=4\pi rk{Q}_{enc}$$ $$E=\frac{2k}{r}\frac{Q_{enc}}{L}$$ $$Q_{enc}=\lambda L+\sigma (2\pi cL)$$ $$E=\frac{2k}{r}\frac{\lambda L+\sigma (2\pi cL)}{L}$$ $$E=\frac{2k}{r}(\lambda +\sigma (2\pi c))$$

Modify For $b\to a$:

$$Q_{enc}=\lambda L$$ $$E=\frac{2k}{r}(\lambda )$$