The Electric Field — Question 10

Question 10

A spatial field gradient determines the change sampled along a path

A local static field is $\mathbf E=(ay)\hat{\mathbf x}+(ax)\hat{\mathbf y}$, with $a=50.0$ N/(C m). A probe is moved along $x=vt$, $y=2vt$, with $v=.100$ m/s. The probe is sufficiently weak to leave the field unchanged.

1. Find the field sampled along the path and its time derivative.

2. At $t=3.00$ s, find the force on a $-2.00$ nC probe and the angle between field and path velocity. Explain why the sampled field changes even though the source is static.

Question 10 - Solution

A spatial field gradient determines the change sampled along a path

(a) Method and calculation

$$\mathbf E(t)=(2avt,avt)=(10t,5t)\,\mathrm{N/C},\quad\frac{d\mathbf E}{dt}=(10,5)\,\mathrm{N/(C\,s)}.$$$$\mathbf E(3)=(30,15)\,\mathrm{N/C},\quad\mathbf F=(-60,-30)\,\mathrm{nN}.$$

(b) Interpretation and further calculation

$$\mathbf v=(v,2v),\quad\cos\theta=\frac{(2,1)\cdot(1,2)}{\sqrt5\sqrt5}=\frac45\Rightarrow\theta\approx36.8699^\circ.$$$$\frac{d\mathbf E}{dt}=(\mathbf v\cdot\nabla)\mathbf E\quad\text{here}.$$Moving through a nonuniform static field changes what the probe samples without requiring any explicit time dependence of the field at a fixed point.

Physical check

A field-line direction and a prescribed probe path need not coincide.