A+1.0μc charge sits at the origin, another +1.0μc sits at x = +0.01m, and another -2.0μc sits at x = +0.02m. calculate the magnitude and direction of the electric field at the point x = +0.2m. (1 μc = 1.0x10^-6c). draw a picture first.

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

i think its c if i'm wrong