Calculating the Potential from the Field — Question 1

Question 1

A vector displacement samples two components of a uniform field

A uniform field is $\mathbf E=(120\hat{\mathbf x}-80\hat{\mathbf y})$ V/m. A is at the origin and B at $(.300,.400)$ m. Set $V_A=25.0$ V.

1. Find $V_B$ by a line integral along the straight segment.

2. Repeat along a horizontal then vertical path and find field work on a 5.00 nC probe.

Question 1 - Solution

A vector displacement samples two components of a uniform field

(a) Method and calculation

$$\Delta V=-\mathbf E\cdot\Delta\mathbf r=-[120(.300)-80(.400)]=-4.00\,\mathrm V.$$$$V_B=25.0-4.00=21.0\,\mathrm V.$$

(b) Interpretation and further calculation

$$\Delta V=-\int_0^{.300}120\,dx-\int_0^{.400}(-80)\,dy=-36+32=-4.00\,\mathrm V.$$$$W_e=-q\Delta V=20.0\,\mathrm{nJ}.$$

Physical check

The vector projection includes cancellation; integrating the field magnitude along the path would give the wrong voltage.