Three charges lie along the x-axis. One positive charge, q1 = 4.80*10^-18 C, is at x = 3.72 m, and another positive charge, q2 = 1.60*10^-19 C, is at the origin.
At what point on the x-axis must a negative charge, q3, be placed so that the resultant force on it is zero?

Answer:

The third charge needs to be placed at [tex]x \approx 0.57\; \rm m[/tex].

Explanation:

Both [tex]q_1[/tex] and [tex]q_2[/tex] would attract [tex]q_3[/tex].

These two electrostatic attractions need to balance one another. Hence, they need to be opposite to one another. Therefore, [tex]q_1[/tex] and [tex]q_2[/tex] need to be on opposite sides of [tex]q_3[/tex]. That is possible only if [tex]q_3 \![/tex] is on the line segment between [tex]q_1 \![/tex] and [tex]q_2 \![/tex].

Assume that [tex]q_3[/tex] is at [tex]x\; \rm m[/tex], where [tex]0 < x < 3.72[/tex] (in other words, [tex]q_3 \![/tex] is on the line segment between [tex]q_1[/tex] and [tex]q_2[/tex], and is [tex]x\; \rm m \![/tex] away from [tex]q_2 \![/tex].)

Let [tex]k[/tex] denote Coulomb's constant.

The magnitude of the electrostatic attraction between [tex]q_1[/tex] and [tex]q_3[/tex] would be:

[tex]\displaystyle \frac{k\cdot q_1 \cdot q_3}{(3.72 - x)^{2}}[/tex].

Similarly, the magnitude of the electrostatic attraction between [tex]q_2[/tex] and [tex]q_3[/tex] would be:

[tex]\displaystyle \frac{k\cdot q_2 \cdot q_3}{x^{2}}[/tex].

The magnitudes of these two electrostatic attractions need to be equal to one another for the resultant electrostatic force on [tex]q_3[/tex] to be [tex]0[/tex]. Equate these two expressions and solve for [tex]x[/tex]:

[tex]\displaystyle \frac{k\cdot q_1 \cdot q_3}{(3.72 - x)^{2}} = \frac{k\cdot q_2 \cdot q_3}{x^{2}}[/tex].

[tex]\displaystyle \frac{q_1}{(3.72 - x)^{2}} = \frac{q_2}{x^{2}}[/tex].

[tex]\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{q_2}{q_1}[/tex].

[tex]\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{q_2}{q_1} = \frac{1}{30}[/tex].

By the assumption that [tex](0 < x < 3.72)[/tex], it should be true that [tex](x > 0)[/tex] and [tex](3.72 - x > 0)[/tex]. Therefore, [tex]\displaystyle \frac{x}{(3.72 - x)} > 0[/tex].

Take the square root of both sides of the equation [tex]\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{1}{30}[/tex].

[tex]\displaystyle \sqrt{\frac{x^2}{(3.72 - x)^{2}}} = \sqrt{\frac{1}{30}}[/tex].

[tex]\displaystyle \frac{x}{3.72 - x} = \frac{1}{\sqrt{30}}[/tex].

[tex]\sqrt{30}\, x = 3.72 - x[/tex].

Therefore:

[tex]\left(1 + \sqrt{30}\right)\, x = 3.72[/tex].

[tex]\displaystyle x = \frac{3.72}{1 + \sqrt{30}} \approx 0.57[/tex].

Hence, [tex]q_3[/tex] should be placed at [tex]x \approx 0.57\; \rm m[/tex].

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