# I tried to solve this problem (Q (a)) but i couldnt find the value of q. any idea how to solve it?

### 1 Answer

- billrussell42Lv 73 months agoFavorite Answer
Forces on q₁ are as follows

F₂ (due to q₂) = kq₁q₂/a²

F₂x = kq₁q₂/a², F₂y = 0

F₃ (due to q₃) = kq₁q₃/a²

F₃x = 0, F₃y = kq₁q₃/a²

F₄ (due to q₄) = kq₁q₄/(√2a)² = kq₁q₄/2a²

F₄x = F₄cos45 = (1/√2)kq₁q₄/2a²

F₄y = F₄sin45 = (1/√2)kq₁q₄/2a²

using q₂ = q₃ = q and q₄ = q₁

1/(2√2) = √2/4

F₂x = kq₁q/a², F₂y = 0

F₃x = 0, F₃y = kq₁q/a²

F₄x = F₄cos45 = (√2)kq₁²/(4a²)

F₄y = F₄sin45 = (√2)kq₁²/(4a²)

add up the x and y components and set equal to zero

Fx = (kq₁q/a²) + (√2)kq₁²/(4a²) = 0

Fy = (kq₁q/a²) + (√2)kq₁²/(4a²) = 0

so one of the forces is opposed to the other, eg, q₁ is negative

(kq₁q/a²) = (√2)kq₁²/(4a²)

q = (√2)q₁/(4) = 0.354q₁

Coulomb's law, force of attraction/repulsion

F = kQ₁Q₂/r²

Q₁ and Q₂ are the charges in coulombs

F is force in newtons

r is separation in meters

k = 8.99e9 Nm²/C²