18.4 Electric field of a point charge

Learning objectives

18.4.1E = Q / (4πε0 r 2) for the electric field strength due to a point charge in• recall and use E = Q / (4πε0 r 2) for the electric field strength due to a point charge in free space

$field magnitude E=|Q|/(4πε0r²), ε0=8.85×10⁻¹² F m⁻¹$

Source charge Q	Field direction at the point
positive	radially away from Q
negative	radially toward Q

The field belongs to the source charge and is independent of any test charge. For a charged spherical conductor and an external point, use the sphere's total charge at its centre and measure r from the centre.

For Q=83 pC and r=0.620 m, E=(83×10⁻¹²)/[4π(8.85×10⁻¹²)(0.620)²]=1.94 N C⁻¹, directed outward because Q is positive.

$For two source contributions, E1/E2=(|Q1|/|Q2|)(r2²/r1²); the common factor 1/(4πε0) cancels.$

Superposition step	Action
1	calculate each field magnitude at the point
2	draw each direction away from + or toward -
3	add vectors; add collinear same directions and subtract opposite directions

Do not insert a test charge into the field formula—that belongs in F=qE. Field strength falls as 1/r², and multiple fields must be added as vectors, not automatically as magnitudes.

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—18.5 Electric potentialRead