Question

A point charge is at the origin. With this point charge as the source point, what is the unit vector$\hat{\mathbf{r}}$in the direction of the field point (a) at x = 0, y = -1.35 m; (b) at x = 12.0 cm, y = 12.0 cm; (c) at x = -1.10 m, y = 2.60 m? Express your results in terms of the unit vectors$\hat{\mathbf{i}}$and$\hat{\mathbf{j}}$.

Short answer

1. $\widehat{r}=-\widehat{j}$
2. $\widehat{r}=\frac{\sqrt{2}}{2}\widehat{i}+\frac{\sqrt{2}}{2}\widehat{j}$
3. $\widehat{r}=-0.390\widehat{i}+0.921\widehat{j}$

Step-by-step solution

Step 1:

As per the formula;

$$\begin{gathered} \widehat{r}=\frac{\overrightarrow{r}}{\left|\overrightarrow{r}\right|} \\ \widehat{r}=\cos \theta \widehat{i}+\sin \theta \widehat{j} \\ \widehat{r}=\frac{0\widehat{i}-1.35\widehat{j}}{\sqrt{0^2+(1.35{)}^2}}=\frac{1}{1.35}(0\widehat{i}-1.35\widehat{j})=-\widehat{j} \end{gathered}$$

Step 2:

$$\begin{gathered} \theta =\tan -1\left(\frac{y}{x}\right)=\tan -1\left(\frac{0.120}{0.120}\right)=\frac{\pi }{4} \\ \widehat{r}=\cos {\left(\frac{\pi }{4}\right)}^{\widehat{i}}+\sin {\left(\frac{\pi }{4}\right)}^{\widehat{j}} \\ \widehat{r}=\frac{\sqrt{2}}{2}\widehat{i}+\frac{\sqrt{2}}{2}\widehat{j} \end{gathered}$$

Step 3:

$$\begin{gathered} \widehat{r}=\frac{-1.10\widehat{i}+2.60\widehat{j}}{\sqrt{(-1.10{)}^2+(2.60{)}^2}} \\ =-0.390\widehat{i}+0.921\widehat{j} \end{gathered}$$