Franco asked in Science & MathematicsPhysics · 10 months ago

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 -.

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2 Answers

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  • 10 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 <<<<<<

  • 10 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

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