Electric Fields in Extra Dimensions

Electricity and Magnetism Level 1

How does the magnitude of an electric field around a point charge scale with increasing radius in d d spatial dimensions? For example, in three spatial dimensions, Coulomb's Law yields E ( r ) ∝ 1 r 2 E(r) \propto \frac{1}{r^2} .

r d + 1 r^{d+1} 1 r d ! / 2 − 1 \frac{1}{r^{d!/2 - 1}} 1 r d − 1 \frac{1}{r^{d-1}} 1 r 2 d − 4 \frac{1}{r^{2d-4}}

Matt DeCross by Matt DeCross , 27, USA

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1 solution

Matt DeCross
Jan 23, 2016

One can use Gauss' Law and spherical symmetry to write:

Q = ∫ ρ d V = ∫ ∇ ⃗ ⋅ E ⃗ d V = ∫ E ⃗ ⋅ d A ⃗ = ∣ E ⃗ ∣ ∫ d A = E ( r ) ⋅ A ( r ) . Q = \int \rho dV = \int \vec{\nabla} \cdot \vec{E} dV = \int \vec{E} \cdot d\vec{A} = |\vec{E}| \int dA = E(r)\cdot A(r).

The volume of the ball in d d dimensions of radius r r scales like the dimension, r d r^d . The surface area of that ball is one dimension lower: r d − 1 r^{d-1} . So the magnitude of the electric field in d d dimensions scales as E ( r ) ∝ 1 r d − 1 E(r) \propto \frac{1}{r^{d-1}} as claimed.

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