Putting it all together 2

Electricity and Magnetism Level 3

By using a sphere centered on the origin one can take advantage of the symmetry of the problem. A point charge at the origin is spherically symmetric - it looks the same no matter which direction you look at the system. Therefore, choosing a surface that is also spherically symmetric means that $|\vec{E}|$ is a constant on the surface. Furthermore, due to spherical symmetry the electric field must point radially outward. Therefore the angle between $\vec{E}$ and $d\vec{A}$ is always zero and the dot product is a constant. Hence Gauss' law

$\int_{S} \vec{E} \cdot \vec{dA}=\frac{Q_{enc}}{\epsilon_0}$

becomes the simpler

$|\vec{E}| \int_{S} dA=\frac{Q_{enc}}{\epsilon_0}$ ,

or, using that the integral of $dA$ is just the surface area,

$|\vec{E}|=\frac{Q_{enc}}{\epsilon_0 A}$ .

If a charge of magnitude $\epsilon_0$ Coulombs is placed at the origin, what is the electric field at the point (1,1,1)?

(4 \pi)^{-1} \hat{r}

(12 \pi)^{-1}

(12 \pi)^{-1} \hat{r}

(4 \pi)^{-1}

1 solution

Muhammad Arifur Rahman
Oct 8, 2015

You just have to use the equation which is given by David Mattingly sir already? As $q=\epsilon_0$ , $|\vec{E}|=\frac{1}{\text{A}}$

Now consider a cube with side length ,a=1m . The length of the diagonal of this cube = radius of the Gaussian sphere $=\sqrt{3}a=\sqrt{3}$

Then $\text{A}=4\pi (\sqrt{3})^2=12\pi$

Finally, $|\vec{E}|=\frac{1}{12\pi}\times \hat{r}=\boxed{(12\pi)^{-1} \hat{r}}$

Putting it all together 2

1 solution

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