Question

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?

Answer 1

Answer:

If the third charge is placed at a distance $b=\sqrt{a^{2} +2ab}$ 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

$F1 = \frac{k2Q^{2} }{(a+b)^{2} }$

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

$F2= \frac{-kQ^{2} }{b^{2} }$

For the third charge in equilibrium,

$\\F1 + F2 = 0\\\\\frac{k2Q^{2} }{(a+b)^{2} } - \frac{kQ^{2} }{(b)^{2} } =0$

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

$\frac{k2Qx^{2}b ^{2}-kQ^{2}(a+b)^{2} }{(a+b)^{2}b^{2} }=0\\k2Qx^{2}b ^{2}-kQ^{2}(a+b)^{2} = 0\\\\kQx^{2}(2b ^{2}-(a+b)^{2}) = 0\\((2b ^{2}-(a+b)^{2}) = 0\\\\(2b ^{2}-a^{2}-2ab-b^{2}=0\\b^{2}-a^{2} -2ab=0\\a^{2} -b^{2} + 2ab =0$

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

$b=\sqrt{a^{2} +2ab}$

So if the third charge is placed at a distance $b=\sqrt{a^{2} +2ab}$ 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