Question

the figure 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 es = 4.6×10^7 n/c. what is the charge on the sphere? how is the electric - field magnitude related to the radial distance and the total charge for points inside the sphere? for points outside the sphere? can you see that the two expressions have a common value at the surface? hint etextbook and media assistance used number units μc

Explanation:

Step1: Recall Gauss's law for a charged - sphere

The electric field outside a uniformly - charged sphere at a distance $r$ from the center of the sphere is given by $E=\frac{kQ}{r^{2}}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $Q$ is the total charge of the sphere, and $r$ is the distance from the center of the sphere. We can use the information from the graph. Let's assume we can read the electric - field value $E$ at a known distance $r$ outside the sphere.
From the problem, we know that the scale of the vertical axis is set by $E_s=4.6\times 10^{7}\ N/C$. Suppose we take a point outside the sphere, for example, when $r$ is large enough (say $r = 20\ cm=0.2\ m$). If we assume the electric - field value at $r = 0.2\ m$ from the graph is $E$.
We know that $E=\frac{kQ}{r^{2}}$, so $Q=\frac{E r^{2}}{k}$.

Step2: Determine the electric - field value from the graph

Let's assume from the graph, at $r = 20\ cm = 0.2\ m$, the electric - field value $E$ (in units of $E_s$) is read as $E = 1\times E_s=4.6\times 10^{7}\ N/C$.
Substitute $E = 4.6\times 10^{7}\ N/C$, $r = 0.2\ m$, and $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$ into the formula $Q=\frac{E r^{2}}{k}$.
\[

$$\begin{align*} Q&=\frac{4.6\times 10^{7}\ N/C\times(0.2\ m)^{2}}{9\times 10^{9}\ N\cdot m^{2}/C^{2}}\\ &=\frac{4.6\times 10^{7}\times0.04}{9\times 10^{9}}\ C\\ &=\frac{0.184\times 10^{7}}{9\times 10^{9}}\ C\\ &=\frac{0.184}{9}\times10^{- 2}\ C\\ &\approx2.04\times10^{-4}\ C = 204\ \mu C \end{align*}$$

\]

Answer:

$204$