Question

At each corner of a square of side l there are point chargesof

magnitude Q, 2Q, 3Q, and 4Q. Determine the force on thecharge
2Q?

Answer 1

F = k Q² / L² (4.82 i ^ - 8.82 j ^ N , F = k Q² / L² 10.05 N , θ = 298⁰

Step-by-step explanation:

To find the electric force we must use Coulomb's law

F = k q₁ q₂ / r₁₂²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², q is the value of the charges and r the distance between them

In our case the total force is

F = F₁₂ + F₃₂ + F₄₂

In the attached we can see the location of the charges. Distance

r₁₂ = L

r₃₂ = L

r₄₂ = √ L² + L² = L √ 2

Let us explicitly write the equation, remember that force is a vector equation,

X axis

Fₓ = F₁₂ +
`F_(42x)`

Y Axis

`F_(y)`= - F₃₂ -
`F_(42y)`

With we have a square the angle of F₄₂ is 45⁰, using trigonometry we can find its components

cos 45 =
`F_(42x)` / F₄₂

sin 45 =
`F_(42y)` / F₄₂

`F_(42x)` = F₄₂ cos 45

`F_(42y)` = F₄₂ sin45

Let's replace

Fₓ = k Q 2Q / L² + k 4Q 2Q / 2L² cos 45

`F_(y)` = - k 3Q 2Q / L² - k 4Q 2Q / 2L² sin 45

Fₓ = k Q² (2 + 4 cos 45) = k Q² 4.82

`F_(y)` = - k Q² (6 + 4 sin 45) = k Q² 8.82

We can give strength in two ways

F = k Q² / L² (4.82 i ^ - 8.82 j ^ N

In the form of magnitude and angle

F = Ra Fₓ² +
`F_(y)` ²

F = kQ² / L² RA (4.82² + 8.82²)

F = k Q² / L² 10.05 N

θ = tan⁻¹
`F_(y)` / Fₓ

θ = tan⁻¹ (-8.82 / 4.82)

θ = -61.34 = 298⁰