Question

Figure 23-55 shows two non-conducting spherical shells fixed in place on an x-axis. Shell 1 has uniform surface charge density $\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\mu C}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$on its outer surface and radius $\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{cm}$, and shell 2 has uniform surface charge density on its outer surface and radius $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$ ; the centers are separated by $\mathbf{L}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$ . Other than at $\mathbf{x}\mathbf{}\mathbf{=}\mathbf{\infty }$, where on the x-axis is the net electric field equal to zero?

Short answer

The net electric field is equal to zero on the x-axis at 3.3 cm .

Step-by-step solution

The given data

1. Shell 1 has a uniform surface charge density ${\mathrm{\sigma }}_1=+4.0\mathrm{\mu C}/{\mathrm{m}}^2$ on its outer surface and radius ${\mathrm{r}}_1=0.50\hspace{0.33em}\mathrm{cm}$ .
2. Shell 2 has a uniform surface charge density ${\mathrm{\sigma }}_2=-2.0\mathrm{\mu C}/{\mathrm{m}}^2$on its outer surface and radius ${\mathrm{r}}_2=2\hspace{0.33em}\mathrm{cm}$ .
3. The centers of the shells are separated by length,$\mathrm{L}=6\hspace{0.33em}\mathrm{cm}$

Understanding the concept of the electric field

Using the concept of the surface density of a charged material in the formula of the electric field on a point due to the charged particle, we can get the required value of the position of the net electric field.

Formulae:

The surface density of a charged material,

$\mathrm{\sigma }=\frac{\mathrm{q}}{\mathrm{A}}$ (1)

The electric field on a particle due to a given charge,

$\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}$ (2)

Calculation of the position on the x-axis at which the net electric field is zero

The point where the individual fields cancel cannot be in the region between the shells since the shells have opposite-signed charges. It cannot be inside the radius R of one of the shells since there is only one field contribution there (which would not be canceled by another field contribution and thus would not lead to zero net fields). We note shell 2 has a greater magnitude of charge ($\left|{\sigma }_2\right|A_2$) than shell 1, which implies the point is not to the right of shell 2 (any such point would always be closer to the larger charge and thus no possibility for cancellation of equal-magnitude fields could occur). Consequently, the point should be in the region to the left of shell 1 (at a distance $r>R_1$from its center); this is where the condition of the net-zero electric field takes place. The given by using equations (1) and (2) as given:

$$\begin{gathered} {\mathrm{E}}_1={\mathrm{E}}_2 \\ \frac{\left|{\mathrm{q}}_1\right|}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}=\frac{\left|{\mathrm{q}}_2\right|}{4{\mathrm{\pi \epsilon }}_0(\mathrm{r}+\mathrm{L}{)}^2} \\ \frac{\left|{\mathrm{\sigma }}_1\right|{\mathrm{A}}_1}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}=\frac{\left|{\mathrm{\sigma }}_2\right|{\mathrm{A}}_2}{4{\mathrm{\pi \epsilon }}_0(\mathrm{r}+\mathrm{L}{)}^2} \\ \mathrm{r}=\frac{\mathrm{L}}{\left({\displaystyle \frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}}\right)\sqrt{\left(\frac{\left|{\mathrm{\sigma }}_2\right|}{{\mathrm{\sigma }}_1}\right)-1}} \\ \mathrm{r}=\left(\frac{6}{\left(\frac{2\hspace{0.33em}\mathrm{m}}{0.5\hspace{0.33em}\mathrm{m}}\right)\sqrt{\left(\frac{2\mathrm{\mu }}{4\mathrm{\mu }}\right)-1}}\right)\mathrm{cm} \\ \mathrm{r}=3.3\mathrm{cm} \end{gathered}$$

This value satisfies the requirement $r>R_1$. Hence, the electric field vanishes at $3.3\hspace{0.33em}\mathrm{cm}$.