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Formulas/physics/Electric Charges Fields/Field due to Point Charge

Field due to Point Charge

Electric field at distance r from a point charge q. Points radially outward for positive q, inward for negative q.
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Derivation

Derivation

Place a point charge qq at the origin. Put a test charge q0q_0 at position r⃗\vec{r}, at distance r=∣r⃗∣r = |\vec{r}|.

By Coulomb's law, the force on q0q_0:

F⃗=14πε0q q0r2r^\vec{F} = \frac{1}{4\pi\varepsilon_0}\frac{q\, q_0}{r^2}\hat{r}

By definition, E⃗=F⃗/q0\vec{E} = \vec{F}/q_0:

E⃗=14πε0qr2r^\vec{E} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{r} E⃗=q4πε0r2r^\boxed{\vec{E} = \frac{q}{4\pi\varepsilon_0 r^2}\hat{r}}

Direction

  • q>0q > 0: field points radially outward from the charge.
  • q<0q < 0: field points radially inward toward the charge.

Dependence on rr

The magnitude E=kq/r2E = kq/r^2 falls off as the inverse square of distance. This is the same rr-dependence as Coulomb's force itself, which is expected — the field is force per unit charge.

Remember
The field of a point charge has spherical symmetry. For a system of charges, the net field is the vector sum of individual fields (superposition).