2 Answers

Melllvar · Answer 1 · 2022-10-11T01:07:39+0000

Answer:

The magnitude of the net electric field is:

$E_(net)=90.37\: N/c$

Step-by-step 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:

|E_(1)|=k(|q_(1)|)/(d_(1)^(2))

Where:

|E_(1)|=(9*10^(9))(|-2.60*10^(-9)|)/(0.538^(2))

$|E_(1)|=80.84\: N/C$

And E2 will be:

$|E_(2)|=k\frac{d_(2){2}}$

|E_(2)|=(9*10^(9))(|-8.30*10^(-9)|)/(1.36^(2))

$|E_(2)|=40.39\: N/C$

Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.

$E_(net)=\sqrt{E_(1)^(2)+E_(2)^(2)}$

$E_(net)=\sqrt{80.84^(2)+40.39^(2)}$

$E_(net)=90.37\: N/c$

I hope it helps you!

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