Question

Two isolated, concentric, conducting spherical shells have radii${\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{\text{m}}$and${\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\text{m}}$ , uniform charges${\mathbf{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu C}$ and${\mathbf{q}}_{\mathbf{2}}\mathbf{=}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu C}$ , and negligible thicknesses. What is the magnitude of the electric field E at radial distance (a)$\mathbf{r}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\text{m}}$ , (b)$\mathit{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{700}\mathbf{\text{m}}$ , and (c)$\mathit{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{\text{m}}$ ? With $\mathit{V}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$at infinity, what is $V$ at (d)localid="1663936396588" $\mathit{r}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\text{m}}$ , (e)localid="1663936405423" $\mathit{r}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\text{m}}$ , (f) localid="1663936413104" $\mathit{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{700}\mathbf{\text{m}}$, (g)localid="1663936424825" $\mathit{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{\text{m}}$ , (h)localid="1663936434140" $\mathit{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{\text{m}}$ , and (i)localid="1663936441939" $\mathit{r}\mathbf{=}\mathbf{0}$? ( j) Sketch localid="1663936449400" $\mathit{E}\left(r\right)$and localid="1663936460268" $\mathbf{V}\left(r\right)$ .

Short answer

a) The magnitude of electric field is $E=1.687\times$${10}^{3}N/C$.

b) The magnitude of electric field is $E=3.67\times$${10}^{4}N/C$.

c) The magnitude of electric field is $E=0$.

d) The value of electric potential is $V\left(r\right)=6.75\text{kV}$.

e) The value of electric potential is $V\left(r\right)=27\text{kV}$.

f) The value of electric potential is $V\left(r\right)=34.7\text{kV}$.

g) The value of electric potential is $V\left(r\right)=45\text{kV}$.

h) The value of potential is $V=4.5\times {10}^3\text{V}$.

g) The value of potential is $V=45\times {10}^3\text{V}$.

Step-by-step solution

Given data:

For the first spherical shell:

The radius, $R_1=0.5\text{m}$

The charge,$q_1=+2.0\mu C=+2.0\times {10}^{-6}C$

For the second spherical shell:

The radius, $R_2=1.00\text{m}$

The charge, $q_2=1.0\mu C=1.0\times {10}^{-6}C$

Understanding the concept:

The electric potential is define by using following formula.

$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_o}\left(\frac{q_1}{R_1}+\frac{q_2}{R_2}\right) \\ =k\left(\frac{q_1}{R_1}+\frac{q_2}{R_2}\right) \end{gathered}$$

Here, $V$ is the electric potential, ${\epsilon }_0$is the permittivity of free space, $q_1$ and $q_2$ are the charges, and $R_1$ and $R_2$ are the distance, $k$ is the Coulomb’s constant having a value

(a) Calculate the magnitude of the electric field E at radial distance r=4.00 m:

At a radial distance,$r=4.0\text{m}$

$r>R_2>R_1$

The electric field is,

$$\begin{gathered} E=k\frac{q_1+q_2}{r^2} \\ =9.0\times {10}^9\times \frac{\left(2.0\times {10}^{-6}+1.0\times {10}^{-6}\right)}{{\left(4.0\right)}^2} \end{gathered}$$

$=1.687\times {10}^{3}N/C$

(b) Calculate the magnitude of the electric field E at radial distance r=7.00 m:

For radius, $r=0.7\text{m}$

$R_2>r>R_1$

The electric field is,

$$\begin{gathered} E=k\frac{q_1}{r^2} \\ =9.0\times {10}^9\times \frac{2.0\times {10}^{-6}}{{\left(0.7\right)}^2} \\ =3.67\times {10}^{4}N/C \end{gathered}$$

Step 5: (c) Calculate the magnitude of the electric field  E at radial distance r=0.200 m:

If $R_2>R_1>r$

The electric field,$E=0$

Because no charge lies inside the shell.

For $r=0.2\text{m}$, $R_2>R_1>r$

Therefore. the electric field, $E=0$

(d) With  V = 0 at infinity, calculate V  at r=4.00 m :

At infinity the potential,$V=0$

If $r=4.0\text{m}$, $r>R_2>R_1$

Thus, the electric potential is,

$$\begin{gathered} V\left(r\right)=k\left(\frac{q_1+q_2}{r}\right) \\ =9.0\times {10}^9\times \left(\frac{1\times {10}^{-6}+2\times {10}^{-6}}{4.0}\right) \\ =6.75\times {10}^3\text{V} \\ =6.75\text{kV} \end{gathered}$$

(e) With  V at infinity, calculate V  at r=1.00 m :

For$r=1.00\text{m}$; $R_2>r>R_1$

The electric potential is,

$$\begin{gathered} V\left(r\right)=k\left(\frac{q_1}{r}+\frac{q_2}{R_2}\right) \\ =9.0\times {10}^9\left(\frac{2.0\times {10}^{-6}}{1.0m}+\frac{1\times {10}^{-6}}{1.0m}\right) \\ =27\times {10}^3\text{V} \\ =27\text{kV} \end{gathered}$$

(f) With  V = 0 at infinity, calculate V  at r=0.700 m :

For$r=0.7\text{m}$; $R_2>r>R_1$

The electric potential is,

$$\begin{gathered} V\left(r\right)=k\left(\frac{q_1}{r}+\frac{q_2}{R_2}\right) \\ =9.0\times {10}^9\left(\frac{2.0\times {10}^6}{0.7}+\frac{1\times {10}^{-6}}{1.0}\right) \\ =34.7\times {10}^3\text{V} \\ =34.7\text{kV} \end{gathered}$$

(g) With V=0  at infinity, calculate V at r=0.500 m :

For$r=0.5\text{m}$; $R_2>r>R_1$

The electric potential is,

$$\begin{gathered} V\left(r\right)=k\left(\frac{q_1}{r}+\frac{q_2}{R_2}\right) \\ =9.0\times {10}^9\left(\frac{2\times {10}^{-6}}{0.5}+\frac{1\times {10}^{-6}}{1.0}\right) \end{gathered}$$

$$\begin{gathered} V\left(r\right)=45\times {10}^3\text{V} \\ =45\text{kV} \end{gathered}$$

(h) With V=0  at infinity, calculate V at r=0.200 m :

For$r=0.2\text{m}$; $R_2>R_1>r$

$$\begin{gathered} V\left(r\right)=k\left(\frac{q_1}{R_1}+\frac{q_2}{R_2}\right) \\ =9.0\times {10}^9\left(\frac{2\times {10}^{-6}}{0.5}+\frac{1\times {10}^{-6}}{1.0}\right) \\ =4.5\times {10}^3\text{V} \end{gathered}$$

(i) With V = 0  at infinity, calculate V at r=0 m :

For$r=0$; $R_2>R_1>r$

$$\begin{gathered} V\left(r\right)=k\left(\frac{q_1}{R_1}+\frac{q_2}{R_2}\right) \\ =9.0\times {10}^9\left(\frac{2\times {10}^{-6}}{0.5}+\frac{1\times {10}^{-6}}{1.0}\right) \\ =45\times {10}^3\text{V} \end{gathered}$$

(j) Sketch E(r)and V(r) :