The Electric Field — Question 2

Question 2

A negative probe reveals the field direction by reversing its force

A test particle has charge $-4.00$ nC and mass $2.00\times10^{-6}$ kg. Its acceleration at one location is $(.0300\hat{\mathbf i}-.0400\hat{\mathbf j})$ m/s$^2$. Electric force is the only force and the source is unchanged by the probe.

1. Find the electric force and the electric field vector.

2. Find the field magnitude and the force on a $+1.50$ nC probe at the same location.

Question 2 - Solution

A negative probe reveals the field direction by reversing its force

(a) Method and calculation

$$\mathbf F=m\mathbf a=(6.00\hat{\mathbf i}-8.00\hat{\mathbf j})\times10^{-8}\,\mathrm N.$$$$\mathbf E=\mathbf F/q=(-15.0\hat{\mathbf i}+20.0\hat{\mathbf j})\,\mathrm{N/C}.$$

(b) Interpretation and further calculation

$$|\mathbf E|=\sqrt{15^2+20^2}=25.0\,\mathrm{N/C}.$$$$\mathbf F\prime=(1.50\times10^{-9})\mathbf E=(-2.25\hat{\mathbf i}+3.00\hat{\mathbf j})\times10^{-8}\,\mathrm N.$$The field belongs to the source configuration; changing the probe changes the force, not the undisturbed field.

Physical check

Divide by signed charge when finding components; divide force magnitude by charge magnitude when finding field magnitude.