Three equal charges 2.8 µc are located in the xy-plane, one at (0 m, 68 m), another at (71 m, 0 m), and the third at (51 m,−49 m). find the magnitude of the electric field at the origin due to these three charges. the value of coulomb’s constant is 8.98755 × 109 n * m^2 /c^2 . answer in units of n/c

E = 8.85 N/C

Explanation:

Look at the attached graphic:

Because all the charges are positive the electric fields leave the charge towards the point P:

The electric field due to each of the charges is calculated by Coulomb's law:

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)

Problem development

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, E₂, E₃: Electric field at point P due to charge q₁, q₂ and q₃ respectively

q = q₁ = q₂ = q₃ = 2.8 µC = 2.8*10⁻⁶ C

K*q = 8.98755 × 10⁹ N*m²/C² * 2.8*10⁻⁶ C = 25.16*10³ N*m²/ C

E₁= k*q/d₁² = (25.16*10³)/(68)² = 5.44 N/C

E₂ = k*q/d₂² = (25.16*10³)/(71)² = 4.99 N/C

E₃ = k*q/d₃² = (25.16*10³)/(70.72)² = 5.03 N/C

Calculation of the electric field components at the origin of x-y coordinates  

Eₓ = E₁ₓ+E₂ₓ+ E₃ₓ

E₁ₓ = 0

E₂x = -5 N/C

β=tan⁻¹(49/51) = 43.85°

E₃ₓ = -5.03*cos(43.85°) = -3.63 N/C

Eₓ= 0 - 5 N/C - 3.63 N/C  = -8.63 N/C

Ey = E₁y + E₂y + E₃y

E₁y= -5.44 N/C

E₂y = 0

E₃y = 5.03 sin(43.85°)= 3.48 N/C

Ey = -5.44 N/C + 0 + 3.48 N/C = -1.96 N/C

Magnitude of the net electric field at the origin due q₁ ,q₂ and q₃

E = 8.85 N/C  

explanation:

the masses of the objects and the distance between them

that tool is called an altimeter which measures your altitude which can be used to tell which atmosphere you are currently in, including the stratosphere.

hope this ! : )