Question

A Gaussian surface in the form of a hemisphere of radius$\mathbf{R}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{68}\mathbf{}\mathbf{cm}$lies in a uniform electric field of magnitude$\mathbf{E}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{\hspace{0.33em}}\mathbf{N}\mathbf{/}\mathbf{C}$. The surface encloses no net charge. At the (flat) base of the surface, the field is perpendicular to the surface and directed into the surface. What is the flux through

(a) the base and

(b) the curved portion of the surface?

Short answer

a) The flux through the base of the surface is $-0.0253\hspace{0.33em}\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$.

b) The flux through the curved portion of the surface is $0.0253\hspace{0.33em}\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$.

Step-by-step solution

The given data

a) The gaussian hemispherical surface of the radius,$\mathrm{R}=5.68\hspace{0.33em}\mathrm{cm}$.

b) The magnitude of the electric field,$E=2.50N/C$.

Understanding the concept of the electric flux

Using the concept of the Gauss flux theorem, we can get the required value of the flux passing through the base area of the surface. Now, using the concept of the net charge to be zero for a Gaussian surface, we can get the electric flux through the curved surface.

Formula:

The electric flux passing through any surface, $\mathrm{\varphi }=\overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{A}}$ (i)

a) Calculation of the flux through the base area of the surface

We choose a coordinate system whose origin is at the center of the flat base, such that the base is in the x-y plane and the rest of the hemisphere is in the$\mathrm{z}>0$ half-space.

Now, using the given data and area of the circular base,$\mathrm{A}={\mathrm{\pi R}}^2$ in equation (i), we can get the electric flux through the base surface as given:

$$\begin{gathered} \mathrm{\varphi }={\mathrm{\pi R}}^2\left(-\hat{\mathrm{k}}\right).\mathrm{E}\left(\hat{\mathrm{k}}\right) \\ =-{\mathrm{\pi R}}^2\mathrm{E} \\ =-\mathrm{\pi }{\left(0.0568\mathrm{m}\right)}^2\left(2.50\mathrm{N}/\mathrm{C}\right) \\ =-0.0253\hspace{0.33em}\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the flux is $-0.0253\hspace{0.33em}\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$.

b) Calculation of the flux through the curved area of the surface

Since the flux through the entire hemisphere is zero, the flux through the curved surface is given as:

$$\begin{gathered} {\mathrm{\varphi }}_{\mathrm{C}}={\mathrm{\varphi }}_{\mathrm{base}} \\ =0.0253\mathrm{N}{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the flux is $0.0253\mathrm{N}{\mathrm{m}}^2/\mathrm{C}$.