Putting it all together 2

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 E|\vec{E}| is a constant on the surface. Furthermore, due to spherical symmetry the electric field must point radially outward. Therefore the angle between E\vec{E} and dAd\vec{A} is always zero and the dot product is a constant. Hence Gauss' law

SEdA=Qencϵ0 \int_{S} \vec{E} \cdot \vec{dA}=\frac{Q_{enc}}{\epsilon_0}

becomes the simpler

ESdA=Qencϵ0|\vec{E}| \int_{S} dA=\frac{Q_{enc}}{\epsilon_0} ,

or, using that the integral of dAdA is just the surface area,

E=Qencϵ0A|\vec{E}|=\frac{Q_{enc}}{\epsilon_0 A}.

If a charge of magnitude ϵ0\epsilon_0 Coulombs is placed at the origin, what is the electric field at the point (1,1,1)?


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