Question

Figure 23-52 gives the magnitude of the electric field inside and outside a sphere with a positive charge distributed uniformly throughout its volume. The scale of the vertical axis is set by ${\mathit{E}}_{\mathbf{s}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }\mathbf{10}\mathit{N}\mathbf{/}\mathit{C}$. What is the charge on the sphere?

Short answer

The charge on the sphere is $2.2\times {10}^{-6}\mathrm{C}$.

Step-by-step solution

The given data

The scale of the vertical axis,${\mathrm{E}}_{\mathrm{S}}=5.0\times {10}^7\mathrm{N}/\mathrm{C}$

Understanding the concept of the electric field

Using the concept of the electric field, we can get the charge on the particle by substituting the given values in the given formula.

Formula:

The electric field at a point due to a charged particle, $\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}$ (1)

Calculation of the charge on the sphere

We determine the (total) charge on the ball by examining the maximum value $({\mathrm{E}}_{\mathrm{S}}=5.0\times {10}^7\mathrm{N}/\mathrm{C})$shown in the graph (which occurs at localid="1661248841771" $\mathrm{r}=0.020\mathrm{m}$). Thus, from, we obtain the charge on the sphere as given:

$$\begin{gathered} \mathrm{q}=\left(4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2\right)\times \mathrm{E} \\ =\frac{{(0.020\mathrm{m})}^2(5.0\times 10{\mathrm{N}}^7/\mathrm{C})}{9\times {10}^9\mathrm{N}·{\mathrm{m}}^{2/}{\mathrm{C}}^2} \\ =2.2\times {10}^{-6}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is $2.2\times {10}^{-6}\mathrm{C}$.