Question

Point charges 1 mC and −2 mC are located at (3, 2, −1) and (−1, −1, 4), respectively. Calculate the electric force on a 10 nC charge located at (0, 3, 1) and the electric field intensity at that point.

Answer 1

See attached handwritten document for answer

Step-by-step explanation:

Answer 2

Step-by-step explanation:

a) You can compute the force by using the expression:

`F=k\frac{q_1q_2}{[(x-x_o)^2+(y-y_o)^2+(z-z_o)^2]^{(1)/(2)}}`

where k=8.98*10^9Nm^2/C^2 and q1, q2 are the charges. By replacing for the forces you obtain:

`F_T=k[((1*10^(-3)C)(10*10^(-9)C))/([(3-0)^2+(2-3)^2+(-1-1)^2])}][(3-0)\hat{i}+(2-3)\hat{j}+(-1-1)\hat{k}]\\\\ \ \ \ \ +k[((-2*10^(-3)C)(10*10^(-9)C))/([(-1-0)^2+(-1-3)^2+(4-1)^2])}][(-1-0)\hat{i}+(-1-3)\hat{j}+(4-1)\hat{k}]\\\\F_T=6.41*10^(-3)N[3i-j-2k]-6.9*10^(-3)N[-1i-4j+3k]\\\\=0.026N\hat{i}-0.034N\hat{j}-0.033\hat{k}`

b)

`E=k[(1*10^(-3)C)/([(3-0)^2+(2-3)^2+(-1-1)^2])]+k[((-2*10^(-3)C))/([(-1-0)^2+(-1-3)^2+(4-1)^2])]\\\\E=641428.5N/C+690769.23N/C=1332197.73N/C`

`E=k[(1*10^(-3)C)/([(3-0)^2+(2-3)^2+(-1-1)^2])][3i-2j-2k]+\\\\k[((-2*10^(-3)C))/([(-1-0)^2+(-1-3)^2+(4-1)^2])][-1i-4j+3k]\\\\E=641428.5N/C[3i-2j-2k]-690769.23N/C[-1i-4j+3k]=2615054.7i+1480219.92j-4973626.23k\\\\|E|=√((E_x)^2+(E_y)^2+(E_z)^2)=5810896.56N/C`

where we you have used that E=kq/r^2