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A -2.37 µC point charge and a 4.74 µC point charge are a distance L apart.
Where should a third point charge be placed so that the electric force on that third charge is zero? (Hint: Solve this problem by first placing the -2.37 µC point charge at the origin and place the 4.74 µC point charge at x = −L.)

Respuesta :

Answer:

[tex]x=2L\pm \sqrt{\frac{(-2L)^2-4\times 1\times(-L^2)}{2\times 1} }[/tex]

Explanation:

Given:

  • charge on first particle, [tex]q_1=-2.37\times 10^{-6}\ C[/tex]
  • charge on the second particle, [tex]q_2=4.74\times 10^{-6}\ C[/tex]
  • distance between the two charges = L

Now the third charge must be placed  on the line joining the two charges at a distance where the intensity of electric field is same for both the charges that point will not lie between the two charges because they are opposite in nature.

[tex]E_1=E_2[/tex]

[tex]\frac{1}{4\pi\epsilon_0} \times \frac{q_1}{x^2} =\frac{1}{4\pi\epsilon_0} \times \frac{q_2}{(L+x)^2}[/tex]

[tex]\frac{2.37}{x^2} =\frac{4.74}{L^2+x^2+2xL}[/tex]

[tex]2x^2=L^2+x^2+2xL[/tex]

[tex]x^2-2L.x-L^2=0[/tex]

[tex]x=2L\pm \sqrt{\frac{(-2L)^2-4\times 1\times(-L^2)}{2\times 1} }[/tex]

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