Answered

A charge [tex]\( q_1 = 6.33 \mu C \)[/tex] is attracted by a force of 0.115 N to a second charge [tex]\( q_2 \)[/tex] that is 1.44 m away. What is the value of [tex]\( q_2 \)[/tex]? Include the sign of the charge ( [tex]\( + \)[/tex] or [tex]\( - \)[/tex] ).

( [tex]\( \mu \)[/tex] stands for micro.)

[tex]\[
[?] \times 10^{-6} C
\][/tex]



Answer :

Sure, here is a step-by-step solution to find the value of the charge [tex]\( q_2 \)[/tex] given the conditions:

1. Identify the quantities given:
- Coulomb's constant, [tex]\( k = 8.9875517873681764 \times 10^9 \, \text{N m}^2/\text{C}^2 \)[/tex].
- Charge [tex]\( q_1 = 6.33 \, \mu\text{C} = 6.33 \times 10^{-6} \, \text{C} \)[/tex].
- Force, [tex]\( F = 0.115 \, \text{N} \)[/tex].
- Distance between charges, [tex]\( r = 1.44 \, \text{m} \)[/tex].

2. Use Coulomb's Law:
[tex]\[ F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \][/tex]

3. Solve for [tex]\( q_2 \)[/tex]:
[tex]\[ q_2 = \frac{F \cdot r^2}{k \cdot |q_1|} \][/tex]

4. Plug in the known values:
[tex]\[ q_2 = \frac{0.115 \, \text{N} \cdot (1.44 \, \text{m})^2}{8.9875517873681764 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 6.33 \times 10^{-6} \, \text{C}} \][/tex]

5. Calculate the value:
[tex]\[ q_2 = \frac{0.115 \cdot 2.0736}{8.9875517873681764 \times 10^9 \cdot 6.33 \times 10^{-6}} \][/tex]

6. Simplify the numerator and denominator separately:
[tex]\[ q_2 = \frac{0.238464}{8.9875517873681764 \times 10^9 \cdot 6.33 \times 10^{-6}} \][/tex]

7. Combine the constants:
[tex]\[ q_2 = \frac{0.238464}{5.68821556884036957772 \times 10^4} \][/tex]

8. Perform the division:
[tex]\[ q_2 = 4.191579509743604 \times 10^{-6} \, \text{C} \][/tex]

9. Determine the sign:
Since the force between the charges is attractive, and [tex]\( q_1 \)[/tex] is positive ([tex]\(6.33 \, \mu C\)[/tex]), [tex]\( q_2 \)[/tex] must be negative.

Therefore, the value of [tex]\( q_2 \)[/tex] is:
[tex]\[ -4.191579509743604 \times 10^{-6} \, \text{C} \][/tex]