Respuesta :
Explanation:
It is mentioned that the magnitude of given charge is [tex]2.40 \times 10^{-6} C[/tex] and side of the square is 1.22 m.
Hence, we will calculate the electric field due to charge at point A as follows.
[tex]E_{a} = k\frac{q}{r^{2}}[/tex]
= [tex]\frac{9 \times 10^{9} \times 2.4 \times 10^{-6}}{(1.22)^{2}}[/tex]
= [tex]14.512 \times 10^{3} N/C[/tex] (acts along y-axis)
Electric field at D due to the charge at C is as follows.
[tex]E_{c} = k\frac{q}{r^{2}}[/tex]
= [tex]\frac{9 \times 10^{9} \times 2.4 \times 10^{-6}}{(1.22)^{2}}[/tex]
= [tex]14.512 \times 10^{3} N/C[/tex] (acts along y-axis)
Electric field at D due to the charge at B is as follows.
[tex]E_{d} = k\frac{q}{r^{2}}[/tex]
= [tex]\frac{9 \times 10^{9} \times 2.4 \times 10^{-6}}{(1.7253)^{2}}[/tex]
= [tex]7.2564 \times 10^{3} N/C[/tex]
Now, we will resolve it into two components.
[tex]E_{bx} = E_{b} Cos 45 = 5.1318 \times 10^{3} N/C[/tex]
[tex]E_{by} = E_{b} Sin 45 = 5.1318 \times 10^{3} N/C[/tex]
Now, the resultant x component is as follows.
[tex]E_{x} = (14.512 + 5.1318) \times 10^{3} = 19.6438 \times 10^{3} N/C[/tex]
Resultant y component is given as follows.
[tex]E_{y} = (14.512 + 5.1318) \times 10^{3} = 19.6438 \times 10^{3} N/C[/tex]
Therefore, resultant electric field at point D is as follows.
[tex]E = \sqrt{E^{2}_{x} + E^{2}_{y}}[/tex]
= [tex]\sqrt{(19.6438 \times 10^{3})^{2} + (19.6438 \times 10^{3})^{2}}[/tex]
= [tex]27.780 \times 10^{3} N/C[/tex]
And, the direction of electric field is
[tex]\theta = tan^{-1} (\frac{E_{y}}{E_{x}})[/tex]
= [tex]tan^{-1} (1)[/tex]
= [tex]45^{o}[/tex]
This means that net electric field makes an angle of [tex]135^{o}[/tex] with positive x-axis in counter clockwise direction.
