Question

A long, non-conducting, solid cylinder of radius 4.0 cmhas a non-uniform volume charge density that is a function of radial distance rfrom the cylinder axis$\mathit{\rho }\mathbf{=}\mathit{A}{\mathit{r}}^{\mathbf{2}}$. For$\mathit{A}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\mu }\mathbf{}\mathbf{C}\mathbf{/}{\mathbf{m}}^{\mathbf{5}}$, what is the magnitude of the electric field at (a) $\mathit{r}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$ and (b)$\mathbf{r}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$?

Short answer

a) The magnitude of the electric field at r=3 cm is 1.9 N/C .

b) The magnitude of the electric field at r=5 cm is 3.6 N/C .

Step-by-step solution

The given data

a) The radius of the solid cylinder, r=4.0 cm

b) Uniform surface charge density $\rho =A{r}^2$where,$A=2.5\mu \mathrm{C}/{\mathrm{m}}^5$

Understanding the concept of the electric field

Using the concept of volume charge density, we can get the total charge on the body. Then, using the concept of the electric flux theorem of the Gauss theorem, we can get the electric field at the required point. Now, for the second case, we needed to calculate the linear charge density of the material. Then, using this, we calculate the electric field at the point.

Formulae:

The volume of the solid cylinder,$V={\mathrm{\pi r}}^2\mathrm{L}$ (1)

The electric flux of a conducting sheet, $\varphi =\left|E\right|\left(2\mathrm{\pi rL}\right)=\frac{{\mathrm{q}}_{\mathrm{enc}}}{{\mathrm{\epsilon }}_0}$ (2)

The electric field of a solid cylinder, $\left|E\right|=\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}$ (3)

a) Calculation of the electric field at r = 3cm

From equation (1), we can get the given equation as follows:

$dV=2\mathrm{\pi rLdr}$

The charge enclosed using the volume charge density and the above equation is given as:

$$\begin{gathered} q_{enc}={\int }_0^{\mathrm{r}}A{r}^{2}2\mathrm{\pi rLdr} \\ =\mathrm{A}2\mathrm{\pi L}\underset{0}{\overset{\mathrm{r}}{\int }}{\mathrm{r}}^3\mathrm{dr} \\ =\frac{\mathrm{\pi }}{2}{\mathrm{ALr}}^4 \end{gathered}$$

Now, using the above equation in equation (ii), we get the electric field is given as:

$$\begin{gathered} \left|E\right|=\frac{A{r}^3}{4{\epsilon }_0} \\ =\frac{\left(2.5\mathrm{\mu C}/{\mathrm{m}}^5\right)\times \left(0.030\mathrm{m}\right)}{4\times \left(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\right)} \\ =1.9\mathrm{N}/\mathrm{C} \end{gathered}$$

.

Hence, the value of the electric field is 1.9 N/C .

b) Calculation of the electric field at r = 5cm

Using equation (a), the value of the linear charge density can be given as:

$$\begin{gathered} \lambda =\frac{1}{L}\underset{0}{\overset{0.04}{\int }}Ar2\mathrm{\pi Ldr} \\ =1.0\times {10}^{-11}\mathrm{C}/\mathrm{m} \end{gathered}$$

Now, the electric field at this point can be given using equation (3) as follows:

$$\begin{gathered} \left|\mathrm{E}\right|=\frac{2\times \left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\times \left(1.0\times {10}^{-11}\mathrm{C}/\mathrm{m}\right)}{{\left(0.05\mathrm{m}\right)}^2} \\ =3.6\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is 3.6 N/C .