Question

In the diagram, q1 = +6.60*10^-9 C and q2 = +3.10*10^-9 C. Find the magnitude of the total electric field at point P. pls help?

Answer 1

A I just answer that question

Answer 2

Answer:

|E(t)| = 1258,46 [N/C]

α = 67,5⁰  (angle with respect x-axis)

Explanation:

E(t)  Electric Field is a vector, so we need to determine module and direction

E(t)  =  E(q₁)  + E (q₂)  Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂  respectively.

E(q₁) = K * q₁/ (d₁)²         K = 9 *10⁹   [N*m²/C²]    d₁ = 0,350 m

E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225      [N*m²/C²] *C/m²

E(q₁) = 484,9 [N/C]

E(q₂) =  9 *10⁹ * 3,1*10⁻⁹ / 0,024025

E(q₂) = 1161,29

Then

|E(t)| = √ |Eq₁|² + |Eq₂|²

|E(t)| = √ ( 484,9)² +( 1161,29)²

|E(t)| = √ 235128 + 1348594,46

|E(t)| = 1258,46 [N/C]

And tanα = 1161,29/484,9        tanα =  2,395      α = 67,5⁰

The angle of the vector electric field with the x-axis