Respuesta :
Answer:
Explanation:
Given
[tex]q_1=+5.9\mu C[/tex]
[tex]q_2=+5.9\mu C[/tex]
[tex]q_3=+2.9\mu C[/tex]
[tex]q_4=-8.2\mu C[/tex]
Electric field is given by
[tex]E=\frac{kq}{r^2}[/tex]
Electric field due to [tex]q_1[/tex] and [tex]q_2[/tex] cancel out each other because they point in opposite direction in x-axis
so net Electric field will be because of [tex]q_3[/tex] and [tex]q_4[/tex] towards negative Y axis
[tex]E_3=\frac{kq_3}{r_3^2}[/tex]
[tex]E_3=\frac{9\times 10^9\times 2.9\times 10^{-6}}{(4.8\times 10^{-2})^2}[/tex]
[tex]E_3=1.132\times 10^7[/tex](away from q_3[/tex])
similarly [tex]E_4=1.55\times 10^{7}[/tex] (towards q_4[/tex])
[tex]E_{net}=E_4-E_3=1.55\times 10^{7}-1.132\times 10^7[/tex]
[tex]E_{net}=0.41\times 10^7 N/C[/tex]
Explanation:
In order to find the net electric field, we have to find the four electric fields and add them as vecotrs.
Hence, formula to calculate the electric field is as follows.
E = [tex]\frac{kq}{r^{2}}[/tex]
For the charge at x = 0.042 m (as 1 m = 100 cm)
So, [tex]E_{1} = \frac{9 \times 10^{9} \times 5.9}{(0.042)^{2}}[/tex]
= [tex]3.01 \times 10^{12}[/tex] N (toward the negative x direction)
For the charge at x = -0.042 m
[tex]E_{2} = \frac{9 \times 10^{9} \times 5.9}{(0.042)^{2}}[/tex]
= [tex]3.01 \times 10^{12}[/tex] N (toward the positive x direction)
For the charge at y = 0.048 m
[tex]E_{3} = \frac{9 \times 10^{9} \times 2.9}{0.048}^{2}}[/tex]
= [tex]11.32 \times 10^{12}[/tex] N (toward the negative y direction)
For the charge at y = 0.069 m
[tex]E_{4} = \frac{9 \times 10^{9} \times 8.2}{0.069)^{2}}[/tex]
= [tex]15.50 \times 10^{12}[/tex] N (toward the positive y)
Therefore, net x is calculated as follows.
The net x = [tex]3.01 \times 10^{12} N - 3.01 \times 10^{12} N[/tex]
= 0 N
The net y = [tex]15.50 \times 10^{12} N - 11.32 \times 10^{12}[/tex] N
= [tex]4.18 \times 10^{12}[/tex] N (positive y)
Total net E is found by the pythagorean theorem as follows.
E = [tex]\sqrt{[(0 N)^{2} + (4.18 \times 10^{12} N)^{2}][/tex]
= [tex]4.18 \times 10^{6}[/tex] N
Thus, we can conclude that the net electric field (magnitude and direction) at the origin is [tex]4.18 \times 10^{6}[/tex] N.
