Respuesta :
Explanation:
Formula for the field at the origin due to the point charge at x-axis is represented as follows.
[tex]E_{x} = \frac{1}{4\pi \epsilon_{o}}(\frac{q_{1}}{r^{2}}[/tex]
Putting the given values into the above formula as follows.
[tex]E_{x} = \frac{1}{4\pi \epsilon_{o}}(\frac{q_{1}}{r^{2}}[/tex]
= [tex]\frac{1}{4 \times 3.14 \times 8.85 \times 10^{-12}}(\frac{20 \times 10^{-9}}{(2)^{2}}[/tex]
= 44.981 N/C
This field is towards negative x-axis as field from positive charge are in a direction outwards from them. Hence, expression for this electric field will be written as follows.
[tex]\vec{E}_{1} = -\vec{E_{x}}\hat{i}[/tex]
= [tex]-44.981\hat{i}[/tex]
Now, field present due to the charge present at negative y-axis field at the origin is as follows.
[tex]E_{-y} = \frac{1}{4 \pi \epsilon_{o}} \times \frac{25 \times 10^{-9}}{9}[/tex]
= [tex]\frac{25 \times 1000}{1000.404}[/tex]
= 24.989 N
From negative charge, the field towards negative y-axis is as follows.
[tex]\hat{E_{2}} = -E_{-y}\hat{j}[/tex]
= [tex]-24.989\hat{j}[/tex]
Now, the net field at the origin is as follows.
[tex]\vec{E_{n}} = \vec{E_{1}} + \vec{E_{2}}[/tex]
= [tex]-44.981\hat{i} - 24.989\hat{j}[/tex]
Hence, the direction of net field is w.r.t to x-axis is as follows.
[tex]\theta = tan^{-1}(\frac{E_{x}}{E_{y}})[/tex]
= [tex]tan^{-1}(\frac{24.989}{44.981})[/tex]
= [tex]29.054^{o}[/tex] w.r.t negative x-axis
Therefore, we can conclude that the direction of the electric field at the origin is negative x-axis.
