Answer:
Epx= - 21.4N/C
Epy= 19.84N/C
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
[tex]r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}[/tex]
[tex]sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246[/tex]
[tex]cos\beta =\frac{1}{\sqrt{1.64} } =0.7808[/tex]
Calculation of the electric field at point P due to q1
Ep₁x=0
[tex]Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}[/tex]
Calculation of the electric field at point P due to q2
[tex]Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}[/tex]
[tex]Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}[/tex]
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
