point charges q1=50 uc and q2=-25 uc are placed 1 m apart. what is the force on a third chare q3=2 uc placed midway between q1 and q2? where must q3 of the preceding problem be placed so that the net force on it is zero?

Respuesta :

Answer:

d = -1 m

The negative sign indicates that the charge from that force of the space of the two spheres.

Explanation:

That is a problem of electric forces, given by Coulomb's law

          F = [tex]k \frac{ q1q2}{r^2}[/tex]

We use that charges of the same sign repel and charges of different signs do not attract, so the net force is

           ∑ = F₁₃ + F₂₃

          F_ {net} = [tex]k \frac{q_1q_3}{r_{13}^2} + k \frac{q_2q_3}{ r_{23}^}[/tex]

a) the charge is placed at the midpoint between the other two

          r₁₃ = r₁₂ = R = ½ m = 0.5m

         F_ {net} =[tex]\frac{k}{R^2 } \ q3 ( q1+q2)[/tex]

calculate us

          F_ {net} = 9 10⁹ / 0.5²   2 10⁻⁶ (50 -25) 10⁻⁶

          F_ {net} = 1,800 N

b) where must be placed q3 so that the force is zero

for this case the charge q3 is outside the spheres

          ∑ F = 0

          F₁₁₃ = F₂₃

          k q_1 / r_{13}² = k q₂ q₃ / r₂₃²

          q₁/ r₁₂²   = q₂ / r₂₃²

suppose the distance

          r₁₂ = d

the he other sphere is

          r₂₃ = d + 1

             

we substitute

          q₃ / d² = q₂ / (d + 1) ²

          (d + 1) ² = q₂ / q₃ d²

           d² (1 - q₂/ q₃) + 2d + 1 = 0

we solve the equation of a second

            d = [-2 + [tex]\sqrt{2^2 - 4 1 ( 1+25/50}[/tex] ] / 2

             d = -2 /2

             d = -1 m

The negative sign indicates that the charge from that force of the space of the two spheres.