Answer to Question #312351 in Electricity and Magnetism for Claire

Question

1. Two equal positive charges q1 q2 = 4.0 μC are located at (1.20 m, 0) and (-1.20 m, 0), = respectively. What are the magnitude and direction of the total electric force that q1 and q2 exerts on a third charge q3= 7.0 μC located at (0, -0.50 m) ?

2. In problem #1, find the total electric field at (0, -0.50 m).

Answer 1

1. First, we find the vectors for each force to add them in total and with this, we will find the force exerted on q₃:

$\vec{r_{13}}=\vec{r_3}-\vec{r_1}=(-1.20m,-0.50m) \\ \vec{r_{23}}=\vec{r_3}-\vec{r_2}=(1.20m,-0.50m) \\ || \vec{r_{23}}||=|| \vec{r_{13}}||=\sqrt{(-0.50\,m)^2+(\plusmn 1.20\,m)^2} \\ || \vec{r_{23}}||=|| \vec{r_{13}}||=r=1.30\,m$

The size of the distance between q₁=q₂ and q₃ is r = 1.30 m. We can find the normal vectors for each distance as

$\widehat{r_{13}}=(-1.20m,-0.50m)/(1.30\,m)=(-\frac{12}{13},-\frac{5}{12}) \\ \widehat{r_{23}}=(1.20m,-0.50m)/(1.30\,m)=(\frac{12}{13},-\frac{5}{13})$

Then we use $F_{electric}=\sum_{i,j} \dfrac{kq_iq_j}{r_{ij}^2}\widehat{r_{ij}}$ to find the total electric force

$F_{electric}=\dfrac{kq_1q_3}{r^2}\widehat{r_{13}}+\dfrac{kq_2q_3}{r^2}\widehat{r_{23}}$

We proceed to substitute and we use the fact that q₁=q₂:

$\vec{F_{electric}}=\dfrac{kq_1q_3}{r^2}\Big[ \widehat{r_{13}}+\widehat{r_{23}} \Big] \\ \vec{F_{electric}}=\dfrac{kq_1q_3}{r^2}\Big[ (-\frac{12}{13},-\frac{5}{13})+(\frac{12}{13},-\frac{5}{13})\Big] \\ \vec{F_{electric}}=\dfrac{kq_1q_3}{r^2}\cdot (0,-\frac{10}{13})=-(\frac{10}{13}\frac{kq_1q_3}{r^2})\widehat{j} \\ \vec{F_{electric}}=\frac{(-10)(9\times10^{9}Nm^2/C^2)(4\times10^{-6}C)(7\times10^{-6}C)}{(13)(1.30\,m)^2}\widehat{j} \\ \therefore \vec{F_{electric}}=-(0.114702\,N)\widehat{j}$

This means that the magnitude of the electric force is F = 0.114702 N and the direction is $-\widehat{j}$

2. Since the electric force and field for a certain charge are related, we know that $\vec{F_{electric}}=q\cdot\vec{ E_{electric}}$ so we use that relationship to find E:

$\vec{ E_{electric}}=\frac{1}{q_3}\vec{ F_{electric}} \\ \vec{ E_{electric}}=(\frac{1}{7\times 10^{-6}C})(-(0.114702\,N)\widehat{j}) \\ \therefore \vec{ E_{electric}}=-(16386\,N/C)\widehat{j}$

In conclusion, the magnitude of the total electric field at (0, -0.50 m) is 16386 N/C.