Question

two particles are separated by 0.38 m and have charges of $-6.25 \times 10^{-9}$ c and $2.91 \times 10^{-9}$ c. use coulomb’s law to predict the force between the particles if the distance is doubled. the equation for coulomb’s law is $f_e = \frac{kq_1q_2}{r^2}$, and the constant, $k$, equals $9.00 \times 10^9$ n·m²/c². \
\
\bigcirc a. $2.83 \times 10^{-7}$ n \
\bigcirc b. $1.13 \times 10^{-6}$ n \
\bigcirc c. $-1.13 \times 10^{-6}$ n \
\bigcirc d. $-2.83 \times 10^{-7}$ n

Explanation:

Step1: Identify given values

Given \( q_1 = -6.25 \times 10^{-9}\, \text{C} \), \( q_2 = 2.91 \times 10^{-9}\, \text{C} \), \( k = 9.00 \times 10^{9}\, \text{N·m}^2/\text{C}^2 \), original distance \( r = 0.38\, \text{m} \), new distance \( r' = 2r = 0.76\, \text{m} \).

Step2: Apply Coulomb's Law formula

Coulomb's Law is \( F_e = \frac{kq_1q_2}{r'^2} \). Substitute the values:
\[

$$\begin{align*} F_e &= \frac{(9.00 \times 10^{9}) \times (-6.25 \times 10^{-9}) \times (2.91 \times 10^{-9})}{(0.76)^2}\\ &= \frac{(9.00 \times -6.25 \times 2.91) \times 10^{9 - 9 - 9}}{0.5776}\\ &= \frac{(-163.3125) \times 10^{-9}}{0.5776}\\ &\approx -2.83 \times 10^{-7}\, \text{N} \end{align*}$$

\]

Answer:

D. \(-2.83 \times 10^{-7}\, \text{N}\)