A charge 2Q is located at the origin while a second charge –Q is located at x = a. Where should a third charge be placed so that the net force on the charge is zero?

Respuesta :

Answer:

If the third charge is placed at a distance [tex]b=\sqrt{a^{2} +2ab}[/tex] the net force on the charge will be zero

Explanation:

Consider the diagram below.

Let us add a new charge Q at a distance b from the origin.

Let the force that the charge 2Q exert on the new charge (Q) be

[tex]F1 = \frac{k2Q^{2} }{(a+b)^{2} }[/tex]

Let the force that charge -Q exerts on the new charge be

[tex]F2= \frac{-kQ^{2} }{b^{2} }[/tex]

For the third charge in equilibrium,

[tex]\\F1 + F2 = 0\\\\\frac{k2Q^{2} }{(a+b)^{2} } - \frac{kQ^{2} }{(b)^{2} } =0[/tex]

Solving this will give us a quadratic equation which will be further simplified.

[tex]\frac{k2Qx^{2}b ^{2}-kQ^{2}(a+b)^{2} }{(a+b)^{2}b^{2} }=0\\k2Qx^{2}b ^{2}-kQ^{2}(a+b)^{2} = 0\\\\kQx^{2}(2b ^{2}-(a+b)^{2}) = 0\\((2b ^{2}-(a+b)^{2}) = 0\\\\(2b ^{2}-a^{2}-2ab-b^{2}=0\\b^{2}-a^{2} -2ab=0\\a^{2} -b^{2} + 2ab =0\\\\\\[/tex]

Making b the subject of the formula, we can get the distance that the third charge will be placed .

[tex]b=\sqrt{a^{2} +2ab}[/tex]

So if the third charge is placed at a distance [tex]b=\sqrt{a^{2} +2ab}[/tex] the net force on the charge will be zero

if real values were plugged in, it will be easier for us to determine the value of b

Ver imagen tochinwachukwu33

The third charge should be placed at a distance [tex]x=\pm \sqrt{2} \,a[/tex] so that the net force on the charge is zero.

The answer can be explained as follows.

Coulomb's Law

  • Let a third charge Q be placed at x = b.

According to Coulomb's law, the Force on the charge Q due to the charge 2Q is given by;

  • [tex]F_1=k\frac{2Q^2}{b^2}[/tex]

Force on the charge Q due to the charge -Q is given by;

  • [tex]F_2=k\frac{-Q^2}{a^2}[/tex]

The net force on Q is given as;

  • [tex]F_{net}=F_1+F_2=0[/tex]
  • [tex]k\frac{2Q^2}{b^2}-k\frac{Q^2}{a^2}=0[/tex]

[tex]\frac{2}{b^2}-\frac{1}{a^2}=0[/tex]

[tex]\implies2a^2-b^2=0[/tex]

  • Therefore,  [tex]b=\pm \sqrt{2} \,a[/tex]

Find out more about Coulomb's Law here:

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