Question

two point - charges (q_1) and (q_2) are held in place 4.50 cm apart. another point - charge (q = 1.75 mu c) of mass 5.00 g is initially located 3.00 cm from each of these charges and released from rest. you observe that the initial acceleration of (q) is (324 m/s^{2}) parallel to the line connecting the two point - charges. find (q_1) and (q_2).

Explanation:

Step1: Identify relevant forces and laws

We use Coulomb's law $F = k\frac{q_1q_2}{r^2}$ and Newton's second - law $F = ma$.

Step2: Analyze the symmetry of the problem

Due to symmetry, the vertical components of the electric forces on charge $Q$ from $q_1$ and $q_2$ cancel out. Only the horizontal components contribute to the acceleration. Let the distance between $q_1$ and $q_2$ be $d = 4.50\ cm=0.045\ m$, and the distance between $Q$ and each of $q_1,q_2$ be $r = 3.00\ cm = 0.03\ m$. The mass of $Q$ is $m=5.00\ g = 0.005\ kg$, the charge of $Q$ is $Q = 1.75\ \mu C=1.75\times 10^{- 6}\ C$, and the acceleration of $Q$ is $a = 324\ m/s^2$.

Step3: Calculate the net horizontal force

The net force on $Q$ is $F = ma$. According to Coulomb's law, the horizontal component of the force exerted by $q_1$ (or $q_2$) on $Q$ is $F_x=k\frac{qQ}{r^2}\cos\theta$, where $\cos\theta=\frac{d/2}{r}$. The net horizontal force $F = 2F_x$. So $ma = 2k\frac{qQ}{r^2}\cos\theta$.

Step4: Solve for $q$ (where $q = q_1=q_2$ due to symmetry)

We know that $\cos\theta=\frac{d/2}{r}=\frac{0.045/2}{0.03}=0.75$. Rearranging the equation $ma = 2k\frac{qQ}{r^2}\cos\theta$ for $q$ gives $q=\frac{ma r^2}{2kQ\cos\theta}$. Substituting the values $k = 9\times 10^{9}\ N\cdot m^2/C^2$, $m = 0.005\ kg$, $a = 324\ m/s^2$, $r = 0.03\ m$, $Q = 1.75\times 10^{-6}\ C$, and $\cos\theta = 0.75$ into the formula:
\[

$$\begin{align*} q&=\frac{0.005\times324\times(0.03)^2}{2\times9\times 10^{9}\times1.75\times 10^{-6}\times0.75}\\ &=\frac{0.005\times324\times0.0009}{2\times9\times 10^{9}\times1.75\times 10^{-6}\times0.75}\\ &=\frac{0.005\times324\times0.0009}{2\times9\times1.75\times0.75\times10^{3}}\\ &=\frac{1.62\times0.0009}{2\times9\times1.75\times0.75\times10^{3}}\\ &=\frac{0.001458}{2\times9\times1.75\times0.75\times10^{3}}\\ &=\frac{0.001458}{2\times9\times1.3125\times10^{3}}\\ &=\frac{0.001458}{23.625\times10^{3}}\\ &=6.17\times 10^{-8}\ C = 61.7\ nC \end{align*}$$

\]

Answer:

$q_1 = q_2=61.7\ nC$