Chapter 22 - Electric Fields - Problems - Page 653: 7

$E = (101,428~N/C)~\hat{j}$

We can find the distance from the center to each charge: $r = \sqrt{(2.5~cm)^2+(2.5~cm)^2}$ $r = 3.54~cm$ $r = 0.0354~m$ We can find the magnitude of the electric field at the center due to $q_1$: $E_1 = \frac{\vert q \vert}{4\pi~\epsilon_0~r^2}$ $E_1 = \frac{(10.0\times 10^{-9}~C)}{(4\pi)(8.854\times 10^{-12}~C^2/N~m^2)~(0.0354~m)^2}$ $E_1 = 71,720~N/C$ Since $q_1$ has a positive charge, the electric field at the center due to $q_1$ points away from $q_1$ We can express this electric field in unit-vector notation: $E_1 = (71,720~cos~45^{\circ}~N/C)~\hat{i}+(-71,720~N/C~sin~45^{\circ}~N/C)~\hat{j}$ $E_1 = (50,714~N/C)~\hat{i}+(-50,714~N/C)~\hat{j}$ We can find the magnitude of the electric field at the center due to $q_2$: $E_2 = \frac{\vert q \vert}{4\pi~\epsilon_0~r^2}$ $E_2 = \frac{(20.0\times 10^{-9}~C)}{(4\pi)(8.854\times 10^{-12}~C^2/N~m^2)~(0.0354~m)^2}$ $E_2 = 143,440~N/C$ Since $q_2$ has a negative charge, the electric field at the center due to $q_2$ points toward $q_2$ We can express this electric field in unit-vector notation: $E_2 = (143,440~cos~45^{\circ}~N/C)~\hat{i}+(143,440~N/C~sin~45^{\circ}~N/C)~\hat{j}$ $E_2 = (101,428~N/C)~\hat{i}+(101,428~N/C)~\hat{j}$ We can find the magnitude of the electric field at the center due to $q_3$: $E_3 = \frac{\vert q \vert}{4\pi~\epsilon_0~r^2}$ $E_3 = \frac{(20.0\times 10^{-9}~C)}{(4\pi)(8.854\times 10^{-12}~C^2/N~m^2)~(0.0354~m)^2}$ $E_3 = 143,440~N/C$ Since $q_3$ has a positive charge, the electric field at the center due to $q_3$ points away from $q_3$ We can express this electric field in unit-vector notation: $E_3 = (-143,440~cos~45^{\circ}~N/C)~\hat{i}+(143,440~N/C~sin~45^{\circ}~N/C)~\hat{j}$ $E_3 = (-101,428~N/C)~\hat{i}+(101,428~N/C)~\hat{j}$ We can find the magnitude of the electric field at the center due to $q_4$: $E_4 = \frac{\vert q \vert}{4\pi~\epsilon_0~r^2}$ $E_4 = \frac{(10.0\times 10^{-9}~C)}{(4\pi)(8.854\times 10^{-12}~C^2/N~m^2)~(0.0354~m)^2}$ $E_4 = 71,720~N/C$ Since $q_4$ has a negative charge, the electric field at the center due to $q_4$ points toward $q_4$ We can express this electric field in unit-vector notation: $E_4 = (-71,720~cos~45^{\circ}~N/C)~\hat{i}+(-71,720~N/C~sin~45^{\circ}~N/C)~\hat{j}$ $E_4 = (-50,714~N/C)~\hat{i}+(-50,714~N/C)~\hat{j}$ We can find the net electric field: $E = E_1+E_2+E_3+E_4$ $E = 0~\hat{i}+(101,428~N/C)~\hat{j}$ $E = (101,428~N/C)~\hat{j}$