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

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.