Question

A point charge q1 = -4.00 nC is at the point x = 0.60 m, y = 0.80 m , and a second point charge q2 = +6.00 nC is at the point x = 0.60 m , y = 0. A. Calculate the magnitude of the net electric field at the origin due to these two point charges.
B. Calculate the direction of the net electric field at the origin due to these two point charges.

Answer 1

Answer:

A)Magnitude of the net electric field at the origin due q₁ and q₂

E₀= 131.6 N/C

B) Direction of the net electric field at the origin due q₁ and q₂

β=3.91°

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)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻⁹ C

Data

q₁ = -4 nC = -4*10⁻⁹ C

q₂ = +6nC =+ 6*10⁻⁹ C

k = 9*10⁹ N*m²/C²

$d_{1} =\sqrt{0.6^{2}+0.8^{2} } =1 m$

d₂ = 0.6 m

Graphic attached

The attached graph shows the field in x=0, y=0, due to the charges q₁ and q₂:

E₁: Total field at point x=0 , y=0 due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge.

E₂: Total field at point  x=0 , y=0  due to charge q₂. As the charge q2 is positive (q₂+) ,the field leaves the charge.

Calculation of the electric field at the origin of x-y coordinates due to the charge q₁

E₁=K*q₁/d₁²=9*10⁹*4*10⁻⁹/1²=36N/C

Components (x-y) of the field due to q1:

E₁x=E₁*cosα = 36*(0.6/1)  = 36*(0.6) = 21.6 N/C

E₁y=E₁*sinα  =  36*(0.8/1) = 36*(0.8) =28.8 N/C

Calculation of the electric field at the origin of x-y coordinates due to the charge q₂

E₂=K*q₂/d₂² = -9*10⁹*6*10⁻⁹/0.6² = -150 N/C

Calculation of the electric field components at the origin of x-y coordinates

E₀ is the electric field at the origin of x-y coordinates (x=0,y=0)

E₀x= E₁x+E₂= 21.6-150 = -128.4 N/C

E₀y= E₁y = 28.8 N/C

A )Magnitude of the net electric field at the origin due q₁ and q₂

$E_{o} =\sqrt{128.4^{2}+28.8^{2} }$

E₀= 131.6 N/C

B) Direction of the net electric field at the origin due q₁ and q₂

$\beta =tanx{-1} \frac{E_{oy} }{Eox}$

$\beta =tan^{-1} \frac{28.8}{128.4}$

β=3.91°