# Help with physics plz.?

In the diagram, q2 is +34.4*10^-6 C, and

q3 is -72.8*10^-6 C. The net force on q2

is 225 N to the right. What is q1?

Include the sign of the charge, + or -.

### 2 Answers

- oldschoolLv 710 months ago
E at q2 = 225/q2 = 225/(34.4e-6) =6,540,697 N/C right = E2

But E2 = k*1e-6*[72.8/0.1²+q1/0.1²]

[6,540,697/(k*1e-6) - 7280]/100 = q1 = -65.5µC <<<<<<

- billrussell42Lv 710 months ago
force on Q₁ due to Q₂

F₂ = k(Q₁)(34.4e-6) / 0.1² directed left assuming Q₁ positive

F₂ = k(Q₁)(34.4e-6) / 0.01

F₂ = k(Q₁)(34.4e-4)

force on Q₁ due to Q₃

F₃ = k(Q₁)(72.8e-6) / 0.2² directed right

F₃ = k(Q₁)(72.8e-6) / 0.04

F₃ = k(Q₁)(72.8e-6)(25)

F₃ = k(Q₁)(18.2e-4)

225 = k(Q₁)(72.8e-6)(25) – k(Q₁)(34.4e-4)

225 = k(Q₁)((18.2e-4) – (34.4e-4))

225 = k(Q₁)(–16.2e-4) = –1.458e7(Q₁)

Q₁ = –0.0000154 C = –15.4 µC

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