Question

An electric field as a function of radial coordinate $r$ has the form $\vec{E} = \alpha \dfrac{e^{-r^2}}{r} \hat{r}$, where $\alpha$ is a constant. Assume that dimensions are appropriately taken care of. The electric flux through a sphere of radius $\sqrt{2}$, centered at the origin, is $\Phi$. What is the value of $\dfrac{\Phi}{2\pi\alpha}$ (rounded off to two decimal places)?

Hint:

For electric flux calculations through a sphere, remember that the field is radially symmetric, and use the appropriate integrals for spherical coordinates.

Answer 1

To calculate the electric flux $\Phi$ through a sphere of radius $\sqrt{2}$, we use the definition of electric flux: \[ \Phi = \int \vec{E} . d\vec{A} = \int \alpha \dfrac{e^{-r^2}}{r} \hat{r} . r^2 d\Omega \hat{r}, \] where $d\Omega$ is the solid angle element. The flux simplifies to: \[ \Phi = \alpha \int_0^{2\pi} \int_0^\pi \frac{e^{-r^2}}{r} r^2 \sin(\theta) d\theta d\phi. \] The integrals over $\theta$ and $\phi$ give: \[ \int_0^{2\pi} d\phi = 2\pi, \int_0^\pi \sin(\theta) d\theta = 2. \] Thus the total flux is: \[ \Phi = 2\pi \alpha \int_0^{\sqrt{2}} e^{-r^2} r dr. \] Substituting limits and solving the integral gives the result: \[ \Phi = 2\pi \alpha \left[ \dfrac{e^{-r^2}}{2} \right]_0^{\sqrt{2}} = 2\pi \alpha \times \dfrac{e^{-2} - 1}{2}. \] This simplifies to: \[ \Phi \approx 2\pi \alpha \times 0.220. \] Thus, $\dfrac{\Phi}{2\pi \alpha} \approx 0.220$, which is approximately $2.20$.