Physic Labs

Electricity and magnetism

Electric field and field lines

An electric field describes the electric force on a test charge; field lines show its direction, with their density indicating relative strength.

Charges establish electric fields in space. At a point, the force per unit positive test charge defines the electric-field vector E⃗\vec E. A field is a vector quantity: contributions from several source charges add vectorially.

E⃗=F⃗q0,E=k∣Q∣r2 (point charge),E⃗total=∑iE⃗i\vec E=\frac{\vec F}{q_0},\qquad E= k\frac{|Q|}{r^2}\ (\text{point charge}),\qquad \vec E_{\rm total}=\sum_i\vec E_i

Definition: Electric field and field lines

E⃗\vec E is measured in N/C, equivalently V/m. Its direction is the force direction on a positive test charge. Field lines are drawn tangent to E⃗\vec E everywhere; they leave positive charges and enter negative charges. Field lines do not cross, since the field has only one direction at any point.

Explore field lines from a point charge and an electric dipole, alongside equipotentials perpendicular to the field.

Field lines and source charges

In a diagram, the conventional number of field lines leaving or entering a charge is proportional to its magnitude. Field lines are a visualization, not material threads. The superposition principle gives the total field by adding the field vectors from each source.

F⃗=qE⃗,E⃗dipole(r⃗)=E⃗+(r⃗)+E⃗−(r⃗)\vec F=q\vec E,\qquad \vec E_{\rm dipole}(\vec r)=\vec E_+(\vec r)+\vec E_-(\vec r)

Example: Force on a test charge

At a point, the electric field is 2.0×1032.0\times10^3 N/C to the right. A charge q=−3.0 μq=-3.0\,\muC is placed there. Find the force magnitude and direction.

Solution

F=∣q∣E=3.0×10−6×2.0×103=6.0×10−3F=|q|E=3.0\times10^{-6}\times2.0\times10^3=6.0\times10^{-3} N. Since qq is negative, the force points opposite the field: to the left.

For a point charge QQ, field magnitude falls as 1/r21/r^2, but its direction depends on the source sign: outward for Q>0Q>0 and toward the source for Q<0Q<0. For example, at a point 0.300.30 m from a +2.0 μ+2.0\,\muC charge, E=k∣Q∣/r2≈2.0×105E=k|Q|/r^2\approx2.0\times10^5 N/C. A −1.0-1.0 nC test charge there feels a force of magnitude 2.0×10−42.0\times10^{-4} N, opposite to the field vector. In symmetric arrangements, add vector components rather than adding magnitudes indiscriminately.

Finding the net electric field

Draw each source field vector at the observation point before calculating. Collinear vectors in the same direction add; opposite vectors subtract, leaving the direction of the larger one. For perpendicular vectors, use the Pythagorean theorem for the resultant magnitude. This procedure lets you find the force on any test charge from F⃗=qE⃗total\vec F=q\vec E_{\rm total} after determining the source field, independently of the test charge.

Quick check

What determines the direction of the electric-field vector at a point?

Can two electrostatic field lines intersect at a point?

References

  1. Young, Freedman (2019). University Physics