Answer:
Epₓ=178.78×10⁶ N/C
Explanation:
Conceptual analysis
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 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
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.
Ep₁ₓ: Total field at point P due to charge q₁ in the x direction.
Ep₂ₓ: Total field at point P due to charge q₂ in the x direction.
Known data
q₁ = -3.6 µC = -3.6×10⁻⁶ C
q₂ = 8.5 µC = 8.5×10⁻⁶ C
k = 8.99*10⁹ N*m²/C²
d = 6.8cm = 6.8*10⁻² m
Calculation of r and β
[tex]r=\sqrt{0.8^2+0.8^2} * 10^{-2}m= \sqrt{1.28} * 10^{-2}m[/tex]
[tex]\beta = tan^{-1} (\frac{0.8}{0.8})= tan^{-1} (1)=45^o[/tex]
Problem development
Epₓ: Total field at point P due to charges q₁ and q₂ in the x direction.
Epₓ = Ep₁ₓ + Ep₂ₓ
Ep₂ₓ = 0
Ep₁ₓ = (-k×q₁×Cosβ)/r²
Epₓ = Ep₁ₓ + 0
Epₓ = Ep₁ₓ
[tex]Ep_x=\frac{8.99*10^9*3.6*10^{-6}*Cos(45)}{(\sqrt{1.28}*10^{-2})^2 } = 178.78*10^6\frac{N}{C} [/tex]
