Answer:
Ep= -10400 N/C
Ep= 10400 N/C In the direction of the (-x )axis
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= 9x10⁹N*m²/C²
q₁ = q₃= -5 nC= -5 *10⁻⁹C
q₂ = -2 nC = -2 *10⁻⁹C
d₁=d₃=d
= 10*10⁻²m = 0.1m
d₂ = 6cm = 6*10⁻²m = 0.06m
Graphic attached
The attached graph shows the field due to the charges:
E₁ : Electric Field at point P due to charge q₁. As the charge q₁ is negative (q₁-) ,the field enters the charge
E₂: Electric Field at point P due to charge q₂. As the charge q₂ is negative (q₂-) ,the field enters the charge
E₃: Electric Field at point P due to charge q₃. As the charge q₃ is negative (q₃-) ,the field enters the charge
Ep: Total field at point P due to charges q₁ , q₂ and q₃ .
Because q₁ = q₃ and d₁ = d₃, then, the magnitude of E₁ is equal to the magnitude of E₃
E₁ = E₃ = k*q/d² = ( 9*10⁹*5*10⁻⁹) /(0.1)² = 4500 N/C
E₂= k*q₂/d₂² = ( 9*10⁹*2*10⁻⁹) /(0.06)² = 5000 N/C
Look at the attached graphic :
Epx : Electric field component at point P in x direction
Epx =E₁x+E₂x+ E₃x
E₁x= E₃x= - 4500 * cosβ =- 4500* (6/10) = - 2700 N/C
E₂x = - 5000 N/C
Epx = E₁x+E₂x+ E₃x = - 2700 N/C - 5000 N/C- 2700 N/C
Epx = - 10400 N/C
Epy : Electric field component at point P in y direction
Epy =E₁y+E₂y+ E₃y
E₁y= 4500 * sinβ = 4500* (8/10) = 3600 N/C
E₃y= - 4500 * sinβ = -4500* (8/10) = -3600 N/C
E₂y = 0
Epy =E₁y+E₂y+ E₃y = 3600 N/C+0-3600 N/C
Epy = 0
Magnitude and direction of the electric field this combination of charges produces at a point P
Ep= Epx = 10400 N/C In the direction of the (-x) axis
