Question

Chargeis uniformly distributed in a sphere of radius R.

(a) What fraction of the charge is contained within the radius is r = R/2.00?

(b) What is the ratio of the electric field magnitude at $\mathbf{r}\mathbf{=}\mathbf{R}\mathbf{/}\mathbf{2}\mathbf{.}\mathbf{00}$to that on the surface of the sphere?

Short answer

a) Fraction of the charge contained within the radius is $r=R/2.00is0.125$ .

b) The ratio of the magnitude of the electric field at $r=R/2.00$to that on the surface of the sphere is $\frac{1}{2}$.

Step-by-step solution

Listing the given quantities

• The radius of the sphere R.
• Charge on the sphere Q.

Understanding the concept of electric field

To explain the electrostatic force between the two charges, we assume that the charges create an electric field around them. The magnitude of electric field E generated by the electric charge q at a distance r is given as,

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

Using the concept of the electric field to determine the magnitude of the electric field.

(a) Calculation of the fraction of electric charge

The given situation is explained in the below diagram:

Since the volume contained within a radius of$R/2$ is one-eighth the volume contained within a radius of R, the charge at is$0<r<R/2isQ/8$ . Therefore, the fraction is,

$\frac{1}{8}=0.125$

Therefore, the fraction of the charge contained within the radius$r=R/2.00$ is 0.125.

(b) Calculation of the ratio of the magnitude of the electric field

The electric field produced at the surface by the charged sphere of radius R and charge Q is,

$E=\frac{Q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2}$

At $r=R/2.00$, the magnitude of the field is: $E_{R/2}$

role="math" localid="1657346499822" $$\begin{gathered} E_{R/2}=\frac{Q/8}{4{\mathrm{\pi \epsilon }}_0{\left(R/2\right)}^2} \\ =\frac{1}{2}\frac{Q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2} \end{gathered}$$

The ratio of the electric field at R/2 and at the surface is,

$$\begin{gathered} \frac{E_{R/2}}{E}=\frac{1}{2}\frac{Q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2} \\ \frac{Q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2} \\ =\frac{1}{2} \end{gathered}$$

Thus, The ratio of the electric field at R/2 and at the surface is$\frac{1}{2}$ .