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A point of charge 4.9 μC is placed at the origin (x1 = 0) of a coordinate system, and another charge of -1.1 μC is placed on the x-axis at x2 = 0.28 m. a) Where on the x-axis can a third charge be placed in meters so that the net force on it is zero?
b) What if both charges are positive; that is, what if the second charge is 1.9 μC?

Respuesta :

Answer:

a)The third charge is placed at a distance of r=0.253 m from the charge of [tex]-1.1\ \rm \mu C[/tex] on right side of it

b) The third charge is placed at a distance of r=0.1762 m from the charge of magnitude [tex]4.9\ \rm \mu C[/tex] on its right side on x axis.

Explanation:

Given:

  • Charge of magnitude [tex]4.9\ \rm \mu C[/tex] is placed at origin.
  • Charge of magnitude [tex]-1.1\ \rm \mu C[/tex]  placed at x=0.28 m

a)Let the The third charge is placed at a distance of r m from the charge of [tex]-1.1\ \rm \mu C[/tex] on right side of it.  When one charge is positive and other charge is negative and net force on third charge is zero

[tex]\dfrac{4.9\times Q}{4\pi \epsilon_0(r+0.28)^2}=\dfrac{1.1\times Q}{4\pi \epsilon_0r^2}\\\\r=0.245\ \rm m[/tex]

b) If both charges are positive

Let The third charge is placed at a distance of r from the charge of magnitude [tex]4.9\ \rm \mu C[/tex] on its right side on x axis. then

[tex]\dfrac{4.9\times Q}{4\pi \epsilon_0(r)^2}=\dfrac{1.9\times Q}{4\pi \epsilon_0(0.28-r)^2}\\\\r=0.1726\ \rm m[/tex]