Question

1. a system consists of two positive point charges, q1 and q2>q1. the total charge of the system is +62.0 μc, and each charge experiences an electrostatic force of magnitude 85.0 n when the separation between them is 0.270 m. find q1 and q2.

Explanation:

Step1: Apply Coulomb's law

According to Coulomb's law $F = k\frac{q_1q_2}{r^2}$, where $F = 85.0\ N$, $k=9\times 10^{9}\ N\cdot m^{2}/C^{2}$, $r = 0.270\ m$. So $85.0=9\times 10^{9}\frac{q_1q_2}{(0.270)^{2}}$, which simplifies to $q_1q_2=\frac{85.0\times(0.270)^{2}}{9\times 10^{9}}=6.885\times 10^{-10}\ C^{2}$.

Step2: Use the total - charge relationship

We know that $q_1 + q_2=62.0\times 10^{- 6}\ C$, so $q_2 = 62.0\times 10^{-6}-q_1$.

Step3: Substitute and solve the quadratic equation

Substitute $q_2 = 62.0\times 10^{-6}-q_1$ into $q_1q_2=6.885\times 10^{-10}$. We get $q_1(62.0\times 10^{-6}-q_1)=6.885\times 10^{-10}$. Rearranging gives $q_1^{2}-62.0\times 10^{-6}q_1 + 6.885\times 10^{-10}=0$.
Using the quadratic formula $q_1=\frac{62.0\times 10^{-6}\pm\sqrt{(62.0\times 10^{-6})^{2}-4\times6.885\times 10^{-10}}}{2}$.
First, calculate $(62.0\times 10^{-6})^{2}-4\times6.885\times 10^{-10}=3.844\times 10^{-9}-2.754\times 10^{-9}=1.09\times 10^{-9}$.
Then $\sqrt{1.09\times 10^{-9}}\approx3.3\times 10^{-5}$.
$q_1=\frac{62.0\times 10^{-6}\pm3.3\times 10^{-5}}{2}$.
We have two solutions for $q_1$:
$q_1=\frac{62.0\times 10^{-6}+ 3.3\times 10^{-5}}{2}=\frac{62.0\times 10^{-6}+33.0\times 10^{-6}}{2}=47.5\times 10^{-6}\ C = 47.5\ \mu C$ or $q_1=\frac{62.0\times 10^{-6}-3.3\times 10^{-5}}{2}=\frac{62.0\times 10^{-6}-33.0\times 10^{-6}}{2}=14.5\times 10^{-6}\ C=14.5\ \mu C$.
Since $q_2>q_1$, when $q_1 = 14.5\ \mu C$, $q_2=62.0\times 10^{-6}-14.5\times 10^{-6}=47.5\ \mu C$.

Answer:

$q_1 = 14.5\ \mu C$, $q_2 = 47.5\ \mu C$