Answer:
The magnitude of the net electric field is:
[tex]E_{net}=90.37\: N/c[/tex]
Explanation:
The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).
On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).
Knowing this, the magnitude of the net electric field will be the E1 + E2.
Let's find first E1 and E2.
The electric field equation is given by:
[tex]|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}[/tex]
Where:
- k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
- q1 is the first charge
- d1 is the distance from q1 to P
[tex]|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}[/tex]
[tex]|E_{1}|=80.84\: N/C[/tex]
And E2 will be:
[tex]|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}[/tex]
[tex]|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}[/tex]
[tex]|E_{2}|=40.39\: N/C[/tex]
Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.
[tex]E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}[/tex]
[tex]E_{net}=\sqrt{80.84^{2}+40.39^{2}}[/tex]
[tex]E_{net}=90.37\: N/c[/tex]
I hope it helps you!
