Question

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 1

Answer:

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 $k$ 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 $0$. Equate these two expressions and solve for $x$:

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

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

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

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

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 $x \approx 0.57\; \rm m$.