Question

two charged dust particles exert a force of 6.0×10^(-2) n on each other. part a what will be the force if they are moved so they are only one - eighth as far apart? express your answer using two significant figures. f = value units

Explanation:

Step1: Recall Coulomb's law

$F = k\frac{q_1q_2}{r^2}$, where $F$ is the force, $k$ is a constant, $q_1$ and $q_2$ are charges and $r$ is the distance between them. Let the initial force be $F_1=k\frac{q_1q_2}{r_1^2}=6.0\times 10^{-2}\ N$ and the new - distance $r_2=\frac{1}{8}r_1$.

Step2: Find the new force formula

The new force $F_2 = k\frac{q_1q_2}{r_2^2}$. Substitute $r_2=\frac{1}{8}r_1$ into the formula for $F_2$: $F_2=k\frac{q_1q_2}{(\frac{1}{8}r_1)^2}=k\frac{q_1q_2}{\frac{1}{64}r_1^2}=64\times k\frac{q_1q_2}{r_1^2}$.

Step3: Calculate the new force

Since $F_1 = k\frac{q_1q_2}{r_1^2}=6.0\times 10^{-2}\ N$, then $F_2 = 64F_1$. So $F_2=64\times6.0\times 10^{-2}\ N = 3.84\ N$. Rounding to two - significant figures, $F_2 = 3.8\ N$.

Answer:

$3.8\ N$