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)?


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