Answer:
Part A) Electric fields at the point due to q₁ and q₂:
E₁ = 33.75*10³ N/C (-j) , E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C
Part B) Net electric field at P (Ep)
Ep= (6.48*10³ (-i)+25.11 10³ (-j) )N/C
Explanation:
Conceptual analysis
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
1nC= 10⁻⁹C
1cm= 10⁻²m
Data
k= 9*10⁹ N*m²/C²
q₁ = -6.00 nC = -6 *10⁻⁹C
q₂ = +3.00 nC = +3*10⁻⁹C
d₁ = 4cm = 4 *10⁻²m
d₂ = 5 *10⁻²m
Part A) Calculation of the electric fields at the point due to q₁ and q₂
Look at the attached graphic:
E₁: Electric Field at point P(0,4) cm due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge
E₂: Electric Field at point P(0,4) cm due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge
E₁ = k*q₁/d₁² = 9*10⁹ *6 *10⁻⁹/ (4 *10⁻²)² = 33.75*10³ N/C
E₂ = k*q₂/d₂²= 9*10⁹ *3*10⁻⁹/(5 *10⁻²)² = 10.8*10³ N/C
E₁ = 33.75*10³ N/C (-j)
E₂x=E₂cosβ = 10.8*(3/5) = 6.48*10³ N/C
E₂y=E₂sinβ = 10.8*(4/5) = 8.64*10³ N/C
E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C
Part B) Calculation of the net electric field at P (Ep)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Ep=Epx (i) + Epy (j)
Epx= E₂x= 6.48*10³ N/C (-i)
Epy= E₁y+E₂y= (33.75*10³ (-j) + 8.64*10³ (+j) ) N/C=25.11 10³ (-j) N/C
Ep= (6.48*10³ (-i)+25.11 10³ (-j) )N/C
Ep= (6.48*10³ (-i)+25.11 10³ (-j) )N/C
