Calculating the Field from the Potential — Question 2

Question 2

A directional voltage slope is smaller than the maximum local slope

At P, $\nabla V=(6\hat{\mathbf x}+8\hat{\mathbf y})$ V/m. A unit direction is $\hat{\mathbf u}=(4\hat{\mathbf x}-3\hat{\mathbf y})/5$.

1. Find the voltage derivative and electric-field component along u.

2. Find the direction and rate of steepest voltage decrease, and the force direction for a negative probe.

Question 2 - Solution

A directional voltage slope is smaller than the maximum local slope

(a) Method and calculation

$$D_{\mathbf u}V=\nabla V\cdot\hat{\mathbf u}=(24-24)/5=0.$$$$E_u=-D_{\mathbf u}V=0,\quad|\mathbf E|=\sqrt{6^2+8^2}=10\,\mathrm{V/m}.$$

(b) Interpretation and further calculation

$$\hat{\mathbf n}_{\mathrm{decrease}}=-(3\hat{\mathbf x}+4\hat{\mathbf y})/5,\quad D_{\mathbf n}V=-10\,\mathrm{V/m}.$$$$q<0:\quad\hat{\mathbf F}=+(3\hat{\mathbf x}+4\hat{\mathbf y})/5.$$

Physical check

Zero directional derivative indicates a tangent direction, not a zero gradient.