Electric Field Lines — Question 1

Question 1

Field-line slope specifies a direction rather than a force magnitude

A uniform planar field is $\mathbf E=(30\hat{\mathbf x}+40\hat{\mathbf y})$ N/C. A field line passes through $(.200,.100)$ m. Field-line arrows follow the force on a positive test charge.

1. Find the line equation and its arrow direction.

2. Find the unit tangent and the direction of initial acceleration of a negative charge released from rest on that line.

Question 1 - Solution

Field-line slope specifies a direction rather than a force magnitude

(a) Method and calculation

$$\frac{dy}{dx}=\frac{E_y}{E_x}=\frac43,\quad y-.100=\frac43(x-.200).$$$$\hat{\mathbf t}=\mathbf E/|\mathbf E|=(3/5,4/5).$$

(b) Interpretation and further calculation

$$\mathbf a=q\mathbf E/m,\quad q<0\Rightarrow\hat{\mathbf a}=(-3/5,-4/5).$$$$\theta_E=\tan^{-1}(4/3)\approx53.1301^\circ.$$The field-line arrow remains unchanged when the test charge sign changes.

Physical check

Arrow orientation is part of the field-line convention; a geometric slope alone also permits the opposite orientation.