Question

Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has magnitude of q2 = 5.00 nC and is located at x = 1.00 m, y = 0.800 m, calculate the x and y components, Ex and Ey, of the electric field, , in component form at the origin, (0,0). The Coulomb Force constant is 1/(4π ε0) = 8.99 × 109 N·m2/C2.

Answer 1

Epx= - 21.4N/C

Epy= 19.84N/C

Step-by-step explanation:

Electric field theory

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

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

`r_(2) =\sqrt{1^(2) +0.8^(2) } =√(1.64)`

`sin\beta =(0.8)/(√(1.64) ) =0.6246`

`cos\beta =(1)/(√(1.64) ) =0.7808`

Calculation of the electric field at point P due to q1

Ep₁x=0

`Ep_(1y) =(k*q_(1) )/(r_(1)^(2)  ) =(8.99*10^(9)*2.9*10^(-9)  )/(0.84^(2) ) =36.95(N)/(C)`

Calculation of the electric field at point P due to q2

`Ep_(2x) =-(k*q_(2) *cos\beta )/(r_(2)^(2)  ) =-(8.99*10^(9)*5*10^(-9) *0.7808 )/((√(1.64))^(2)  ) =-21.4(N)/(C)`

`Ep_(2y) =-(k*q_(2) *sin\beta )/(r_(2)^(2)  ) =-(8.99*10^(9)*5*10^(-9) *0.6242 )/((√(1.64))^(2)  ) =-17.11(N)/(C)`

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C