Question

In a rectangular coordinate system, a positive point charge q = 6.50 nC is placed at the point x=0.190 m ,y=0, and an identical point charge is placed at x= -0.190 m , y=0. Find the x and y components and the magnitude and direction of the electric field at the following points. A) Find the x and y components of the electric field at the origin.

B) Find the magnitude of the electric field at the origin.

Answer 1

A) x and y components of the electric field (Ep) at the origin.

Epx = -1620.5 N/C

Epy = -1620.5 N/C

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field (Ep) at the origin.

Ep= 2291.7 N/C

Step-by-step explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

Data

K= 9x10⁹N*m²/C²

q₁ = q₂= +6.5nC=+6.5 *10⁻⁹C

d₁=d₂=0.190m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁ : Electric Field at point P (x=0, y=0) due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge.

Ep₂: Electric Field at point P (x=0, y=0) due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge

Ep: Total field at point P due to charges q₁ and q₂.

Because q₁ = q₂ and d₁ = d₂, then, the magnitude of Ep₁ is equal to the magnitude of Ep₂

Ep₁ = Ep₂ = k*q/d² = 9*10⁹*6.5*10⁻⁹/0.190m² = 1620.5 N/C

Look at the attached graphic :

Epx = Ep₁= -1620.5 N/C

Epy = Ep₂= -1620.5 N/C

A) x and y components of the electric field (Ep) at the origin.

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field (Ep) at the origin.

`E_(p) =\sqrt{1620.5^(2)+1620.5^(2) } = 2291.7 (N)/(C)`