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).
"\\vec{r_{13}}=\\vec{r_3}-\\vec{r_1}=(-1.20m,-0.50m)\n\\\\ \\vec{r_{23}}=\\vec{r_3}-\\vec{r_2}=(1.20m,-0.50m)\n\\\\ || \\vec{r_{23}}||=|| \\vec{r_{13}}||=\\sqrt{(-0.50\\,m)^2+(\\plusmn 1.20\\,m)^2}\n\\\\ || \\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})\n\\\\ \\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]\n\\\\ \\vec{F_{electric}}=\\dfrac{kq_1q_3}{r^2}\\Big[ (-\\frac{12}{13},-\\frac{5}{13})+(\\frac{12}{13},-\\frac{5}{13})\\Big]\n\\\\ \\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}\n\\\\ \\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}\n\\\\ \\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}}\n\\\\ \\vec{ E_{electric}}=(\\frac{1}{7\\times 10^{-6}C})(-(0.114702\\,N)\\widehat{j})\n\\\\ \\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.
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