A)Ep=-81.3N/C :Electric field at the point x = +0.2
Ep Magnitude =81.3N/C
Direction of the electric field ( Ep): -x
B)Graphic attached
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
Graphic attached
The attached graph shows the field due to the charges:
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
E₁: Electric Field at point Xp=0.2 m due to charge q₁. As the charge q1 is positive (q₁+) ,the field leaves the charge.
E₁: Electric Field at point Xp=0.2 m due to charge q₂. As the charge q1 is positive (q₂+) ,the field leaves the charge
E₃: Electric Field at point Xp=0.2 m due to charge q₃. As the charge q₃ is negative (q₃-), the field enters the charge.
Equivalence
+1.0μC=1* 10⁶C
Data
q₁=+1.0μC=1* 10⁶C
q₂=+1.0μC=1* 10⁶C
q₃=-2.0μC=-2* 10⁶C
Xp=0.2m
x₁=0
x₂=0.01 m
x₃=0.02m
Calculation of the distances of the charges to the point P
d= Xp-x
d₁=Xp-x₁= 0.2-0= 0.2m
d₂=Xp-x₂=0.2-0.01= 0.19m
d₃=Xp-x₃=0.2-002= 0.18m
Calculation of electric fields due to charges q1, q2 and q3 at point P
E₁=k*q₁/d₁²=9*10⁹*1*10⁻⁶/0.2²=225*10³N/C
E₂=k*q₂/d₂²=9*10⁹*1*10⁻⁶/0.19²=249.3*10³N/C
E₃=-k*q₃/d₃²=9*10⁹*2*10⁻⁶/0.18²=-555.6*10³N/C
Calculation of electric field at point P due to charges q₁, q₂ and q₃
To calculate Ep, the electric fields E₁,E₂ and E₃ are added algebraically:
Ep=E₁+E₂ +E₃
Ep=(225*10³+249.3*10³ -555.6*10³)N/C
Ep=-81.3N/C
Ep Magnitude =81.3N/C in -x direction
