Electric Charge — Question 6

Question 6

Reversing the test charge reverses every force contribution

At a fixed location, a small test object carrying $q=+2.0$ nC experiences forces $\mathbf F_A=(3\hat{\mathbf i}+4\hat{\mathbf j})$ mN and $\mathbf F_B=(-1\hat{\mathbf i}+2\hat{\mathbf j})$ mN from two unchanged sources. Use the supplied rule that each force is proportional to the test charge when the source arrangement stays fixed.

1. Find the original resultant and its magnitude.

2. Replace the test charge by $-3.0$ nC without disturbing the sources. Find both new force components of the resultant and its direction relative to the original.

Question 6 - Solution

Reversing the test charge reverses every force contribution

(a) Method and calculation

$$\mathbf F=(3-1)\hat{\mathbf i}+(4+2)\hat{\mathbf j}=(2\hat{\mathbf i}+6\hat{\mathbf j})\,\mathrm{mN}.$$$$|\mathbf F|=\sqrt{2^2+6^2}=\sqrt{40}\approx6.325\,\mathrm{mN}.$$

(b) Interpretation and further calculation

$$\lambda=q\prime/q=-3/2,\quad\mathbf F\prime=\lambda\mathbf F=(-3\hat{\mathbf i}-9\hat{\mathbf j})\,\mathrm{mN}.$$$$|\mathbf F\prime|=1.5\sqrt{40}\approx9.487\,\mathrm{mN}.$$Both components reverse, so the direction rotates by $180^\circ$. Adding vector magnitudes would miss cancellation in the horizontal direction.

Physical check

The scaling assumes the source charges and geometry stay fixed; charge-induced rearrangement would need a different model.