Physics, 18.03.2021 02:20 kenken2583

Three charges lie along the x-axis. One positive charge, q1 = 4.8010^-18 C, is at x = 3.72 m, and another positive charge, q2 = 1.6010^-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?

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Answer from: cuppykittyy

The third charge needs to be placed at $x \approx 0.57\; \rm m$ .

Explanation:

Both q_1 and q_2 would attract q_3 .

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

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

Let denote Coulomb's constant.

The magnitude of the electrostatic attraction between q_1 and q_3 would be:

$\displaystyle \frac{k\cdot q_1 \cdot q_3}{(3.72 - x)^{2}}$ .

Similarly, the magnitude of the electrostatic attraction between q_2 and q_3 would be:

$\displaystyle \frac{k\cdot q_2 \cdot q_3}{x^{2}}$ .

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

$\displaystyle \frac{k\cdot q_1 \cdot q_3}{(3.72 - x)^{2}} = \frac{k\cdot q_2 \cdot q_3}{x^{2}}$ .

$\displaystyle \frac{q_1}{(3.72 - x)^{2}} = \frac{q_2}{x^{2}}$ .

$\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{q_2}{q_1}$ .

$\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{q_2}{q_1} = \frac{1}{30}$ .

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

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

$\displaystyle \sqrt{\frac{x^2}{(3.72 - x)^{2}}} = \sqrt{\frac{1}{30}}$ .

$\displaystyle \frac{x}{3.72 - x} = \frac{1}{\sqrt{30}}$ .

$\sqrt{30}\, x = 3.72 - x$ .

Therefore:

$\left(1 + \sqrt{30}\right)\, x = 3.72$ .

$\displaystyle x = \frac{3.72}{1 + \sqrt{30}} \approx 0.57$ .

Hence, q_3 should be placed at $x \approx 0.57\; \rm m$ .

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Three charges lie along the x-axis. One positive charge, q1 = 4.80*10^-18 C, is at x = 3.72 m, and a...

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