Electric Field Lines — Question 10

Question 10

A curved line supplies a local tangent, not a force toward its center of curvature

At P an oriented field line has tangent direction proportional to $(-1,2)$ and field magnitude 90.0 N/C. A particle with charge $-3.00$ nC is placed at P. The drawn curve bends, but its curvature radius is not specified.

1. Find the field and force vectors.

2. Find the force component along a unit normal $(2,1)/\sqrt5$ to the field line. Explain why curvature alone does not create a normal electric force.

Question 10 - Solution

A curved line supplies a local tangent, not a force toward its center of curvature

(a) Method and calculation

$$\hat{\mathbf t}=(-1,2)/\sqrt5,\quad\mathbf E=\frac{90}{\sqrt5}(-1,2)\,\mathrm{N/C}.$$$$\mathbf F=\frac{270}{\sqrt5}(1,-2)\,\mathrm{nN}\approx(120.748,-241.495)\,\mathrm{nN}.$$

(b) Interpretation and further calculation

$$F_n=\mathbf F\cdot\frac{(2,1)}{\sqrt5}=\frac{270}{5}(2-2)=0.$$$$|\mathbf F|=270\,\mathrm{nN}.$$A field line is a directional construction, not a wire constraining motion; its curvature does not imply centripetal acceleration of a freely moving particle.

Physical check

The actual acceleration is parallel or antiparallel to the local tangent, according to the charge sign.