The net electric field is the vector sum of the components of the electric
field produced by the two charges.
The values of the magnitude and direction of the net electric field at the origin (approximate values) are;
- 12.6 ° above the negative x–axis
<h3>How are the net electric field magnitude and direction calculated?</h3>
The possible questions based on a similar question posted online are;
(a) The net electric field at the origin.
The electric field due to charge q₁ is given as follows;
Which gives;
Which gives;
Therefore;
![\vec {E} = \left[ 21.6 \, N/C - 150 \, N/C \right] \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)](https://tex.z-dn.net/?f=%5Cvec%20%20%7BE%7D%20%3D%20%5Cleft%5B%2021.6%20%5C%2C%20N%2FC%20-%20150%20%5C%2C%20N%2FC%20%5Cright%5D%20%5Cleft%28%20%5Chat%20x%20%5Cright%29%20%2B%20%5Cleft%2828.8%20%5C%2C%20N%2FC%20%5Cright%29%20%5Cleft%28%20%5Chat%20y%20%5Cright%29)
The magnitude of the net electric field is therefore;
E = 
≈ 131.6
- The magnitude of the net electric field at the origin is E ≈<u> 131.6 N/C</u>
(b) The direction of the net electric field at the origin.
Learn more about electric field strength here:
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Complete question:
<em>What values of c and d would make the following expression represent a real number? i(2 + 3i)(c + di)
</em>
<em>choices: A)c = 2, d = 3 B)c = –2, d = 3 C)c = 3, d = –2 D)c = –3, d = –2</em>
<em />
<em>Option D</em><em> is correct. </em> The value of c and d for the expression to be real are -3 and -2 respectively
Given the expression
Expand the expression
Collect the like terms
For the resulting function to be real, then the imaginary part must be zero i.e.
If d = -2, then;
Hence the value of c and d for the expression to be real are -3 and -2 respectively.
Learn more here: brainly.com/question/4244460
