Answer:
Ep= 226.68*e*10¹⁸ N/C in direction (+x)
Explanation:
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
1µm= 10⁻⁶m
Data
k= 9*10⁹ N*m²/C²
q₁= e
q₂= -e
d₁=d₂=d
[tex]d=\sqrt{6^{2}+5^{2} } *10^{-6} m[/tex]
d=7,81µm=7,81*10⁻⁶m
Calculation of the electric fields at the point due to q₁ and q₂
Look at the attached graphic:
E₁: Electric Field at point P due to charge q₁. As the charge q₁ is positive (e+) ,the field leaves the charge
E₂: Electric Field at point due to charge q₂. As the charge q₂ is negative (e-), the field enters the charge
Since the magnitude of q₁ is equal to the magnitude of q₂ and the distance of q₁ to the point P is equal to the distance of q₂ to the point P, then, the magnitude of E₁ is equal to the magnitude of E₂:
[tex]E=E_{1} =E_{2} =\frac{k*q}{d^{2} }[/tex]
E=9*10⁹ *e/ (7.81 *10⁻⁶)² =147.5*e*10¹⁸ N/C
E₁x=E₂x=Ecosβ=147.5*e*10¹⁸ *(6/7.81)= +113.34*e*10¹⁸ N/C
E₁y=Esinβ=147.5*e*10¹⁸ *(5/7.81)= +94.43*e*10¹⁸ N/C
E₂y= -Esinβ=-147.5*e*10¹⁸ *(5/7.81)= -94.43*e*10¹⁸ N/C
Components x-y of the electric field at point P due to q₁ and q₂
Epx = E₁x+E₂x =+113.34*e*10¹⁸ +113.34*e*10¹⁸ =+ 226.68*e*10¹⁸ N/C
Epy = E₁y+E₂y =+94.43*e*10¹⁸ - 94.43*e*10¹⁸ = 0
Magnitude and direction of the net electric field at P
Ep=Epx=+ 226.68*e*10¹⁸ N/C in direction (+x)
