Answer:
Epx = -0.81*10⁶ N/C
Epy = 5.79*10⁶ N/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)
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
1µC= 10⁻⁶ C
1cm= 10⁻² m
Graphic attached
The attached graph shows the field due to the charges:
Ep₁: Total field at point P due to charge q₁. As the charge is negative, the field enters the charge.
Ep₂: Total field at point P due to charge q₂. As the charge is positive ,the field leaves the charge.
Known data
q₁ = -2.5 µC = -2.5*10⁻⁶ C
q₂ = 7,2 µC = 7,2*10⁻⁶ C
k = 8.99*10⁹ N*m²/C²
d = 9.9cm = 9.9*10⁻² m
θ = 45°
sinθ = cosθ = [tex]\frac{\sqrt{2} }{2}[/tex]
r calculation
[tex]r=\sqrt{(9.9*10^{-2})^2+(9.9*10^{-2})^2}=0.14m[/tex]
Calculation of the electric field at point P due to q₁
Ep1x = (-k*q₁*cosθ)/r²
[tex]Ep_{1x}= -\frac{8.99*10^9*2.5*10^{-6}\frac{\sqrt{2}}{2}}{0.14^2}=-0.81*10^6 \frac{N}{C}[/tex]
Ep1y = (-k*q₁*sinθ)/r²
[tex]Ep_{1y}= -\frac{8.99*10^9*2.5*10^{-6}\frac{\sqrt{2}}{2}}{0.14^2}=-0.81*10^6 \frac{N}{C}[/tex]
Calculation of the electric field at point P due to q₂
Ep2x=0
Ep2y=k*q₁/r²=(8.99*10⁹*7.2 *10⁻⁶)/
(9.9*10⁻²)² = 6.6*10⁶ N/C
Calculation of the electric field at point P due to q₁ and q₂
Epx = Ep1x + Ep2x = -0.81*10⁶ + 0 = -0.81*10⁶ N/C
Epy = Ep1y+ Ep2y= -0.81*10⁶ + 6.6*10⁶ = 5.79*10⁶ N/C
