Question

Figure 23-36 shows two non-conducting spherical shells fixed in place. Shell 1 has uniform surface charge density$\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\mu C}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$on its outer surface and radius 3.0cm; shell 2 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 2.0 cm ; the shell centers are separated by L = 10cm. In unit-vector notation, what is the net electric field at x= 2.0 cm ?

Short answer

The net electric field at x = 2 cm is$(-2.8\times {10}^4\mathrm{N}/\mathrm{C})\hat{\mathrm{j}}$

Step-by-step solution

The given data

Shell 1:

Uniform charge density$\sigma =+6.0\mathrm{\mu C}/{\mathrm{m}}^2$

Outer surface radius, ${\mathrm{r}}_0=3.0\mathrm{cm}$

Shell 2:

Uniform charge density localid="1657344798468" $\sigma =+4.0\mathrm{\mu C}/{\mathrm{m}}^2$

Outer surface radius,${\mathrm{r}}_0=2.0\mathrm{cm}$

The centers of the shells are separated by L = 10 cm, at x = 2 cm

Understanding the concept of Gauss law-planar symmetry

Using the concept of the electric field at a point at a distance due to another charge, we can get the required value of the net field with its direction.

Formula:

The electric field at a point, $\mathrm{e}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\mathrm{in}\mathrm{the}\mathrm{dirction}\mathrm{of}\mathrm{r}$ (1)

Calculation of the net electric field

We note that only the smaller shell contributes a (nonzero) field at the designated point since the point is inside the radius of the large sphere (and E = 0 inside of a spherical charge) and the field points toward the x-direction. Thus, the net electric field at is given using equation (1) as follows:

$$\begin{gathered} \overrightarrow{\mathrm{E}}=\mathrm{E}\left(-\hat{\mathrm{j}}\right) \\ =-\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\left(\hat{\mathrm{j}}\right) \\ =-\frac{4{\mathrm{\pi R}}^2{\mathrm{\sigma }}_2}{4{\mathrm{\pi \epsilon }}_0{(\mathrm{L}-\mathrm{x})}^2}\hat{\mathrm{j}} \\ =-\frac{{\mathrm{R}}^2{\mathrm{\sigma }}_2}{{\mathrm{\epsilon }}_0{(\mathrm{L}-\mathrm{x})}^2}\hat{\mathrm{j}} \\ \overrightarrow{\mathrm{E}}=\frac{{(0.020\mathrm{m})}^2(4.0\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2)}{(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2){(0.10\mathrm{m}-0.020\mathrm{m})}^2}\hat{\mathrm{j}} \\ =(-2.8\times {10}^4\mathrm{N}/\mathrm{C})\hat{\mathrm{j}} \end{gathered}$$

Hence, the value of the field is $(-2.8\times {10}^4\mathrm{N}/\mathrm{C})\hat{\mathrm{j}}$