1) Coulomb’s law is F = (kq_{1}q_{2}) / (r ^ 2) where k = 9 * 10 ^ 9 in Sl units. Suppose that q * 1 = 6mu*C and q * 2 = 4mu*C and r = 25 cm, calc. the force between these particles. 2) The proton and electron in a H atom both have a charge of 1.6 X 10-19 C They are r = 5 * 10 ^ – 10 meters apart. Calc. the force between them 3) If two charged pith balls have a force of F due to electrostatic attraction, what is the final force on them if the distance between them is doubled? (a) 2F (b) 4F (c) F/2 (d) F/4 4) A charge – 2 C is placed at the origin of the xy plane. A charge of 3 C is placed at x = 25 cm. A final charge of -8 C is placed at x = 50 cm. Calc. the force between the – 2 C and 3 C charges (this force will be in the positive x direction or +i) 5) In problem 4, calc. the force between the – 2 Cand-8 C charges. (this force will be in the negative x direction or – i 6) To get the resultant force on the -2 C charge, add the two force vectors you calculated in problems 4 and 5.
The Correct Answer and Explanation is :
Solutions:
1. Coulomb’s Law Calculation
Given:
- ( q_1 = 6 \, \mu C = 6 \times 10^{-6} \, C )
- ( q_2 = 4 \, \mu C = 4 \times 10^{-6} \, C )
- ( r = 25 \, cm = 0.25 \, m )
- ( k = 9 \times 10^9 \, N \cdot m^2 / C^2 )
[
F = \frac{k \cdot q_1 \cdot q_2}{r^2}
]
[
F = \frac{(9 \times 10^9) \cdot (6 \times 10^{-6}) \cdot (4 \times 10^{-6})}{(0.25)^2}
]
[
F = \frac{(9 \times 10^9) \cdot 24 \times 10^{-12}}{0.0625}
]
[
F = \frac{216 \times 10^{-3}}{0.0625} = 3.456 \, N
]
2. Force Between Proton and Electron
Given:
- ( q_1 = q_2 = 1.6 \times 10^{-19} \, C )
- ( r = 5 \times 10^{-10} \, m )
- ( k = 9 \times 10^9 \, N \cdot m^2 / C^2 )
[
F = \frac{k \cdot q_1 \cdot q_2}{r^2}
]
[
F = \frac{(9 \times 10^9) \cdot (1.6 \times 10^{-19})^2}{(5 \times 10^{-10})^2}
]
[
F = \frac{(9 \times 10^9) \cdot (2.56 \times 10^{-38})}{25 \times 10^{-20}}
]
[
F = \frac{2.304 \times 10^{-28}}{25 \times 10^{-20}} = 9.216 \times 10^{-9} \, N
]
3. Final Force When Distance Doubles
Coulomb’s law states force is inversely proportional to ( r^2 ). If ( r ) doubles, the force becomes:
[
F’ = \frac{F}{2^2} = \frac{F}{4}
]
Answer: ( (d) \, F/4 )
4. Force Between (-2 \, C) and (3 \, C)
Given:
- ( q_1 = -2 \, C ), ( q_2 = 3 \, C )
- ( r = 25 \, cm = 0.25 \, m )
- ( k = 9 \times 10^9 \, N \cdot m^2 / C^2 )
[
F = \frac{k \cdot q_1 \cdot q_2}{r^2}
]
[
F = \frac{(9 \times 10^9) \cdot (-2) \cdot (3)}{(0.25)^2}
]
[
F = \frac{-54 \times 10^9}{0.0625} = -864 \, N \, (\text{positive x-direction, } +\hat{i})
]
5. Force Between (-2 \, C) and (-8 \, C)
Given:
- ( q_1 = -2 \, C ), ( q_3 = -8 \, C )
- ( r = 50 \, cm = 0.5 \, m )
[
F = \frac{k \cdot q_1 \cdot q_3}{r^2}
]
[
F = \frac{(9 \times 10^9) \cdot (-2) \cdot (-8)}{(0.5)^2}
]
[
F = \frac{144 \times 10^9}{0.25} = 576 \, N \, (\text{negative x-direction, } -\hat{i})
]
6. Resultant Force
Net force on (-2 \, C) charge:
[
F_{\text{net}} = F_{14} + F_{15}
]
[
F_{\text{net}} = 864 \, N \, (+\hat{i}) – 576 \, N \, (-\hat{i}) = 288 \, N \, (+\hat{i})
]
Explanation
Coulomb’s law calculates the electrostatic force between charged particles based on their charges, separation distance, and a constant ( k ). Problems 1 and 2 demonstrate simple applications, while problem 3 reinforces the inverse square law of distance on force. Problems 4–6 combine vector addition for charges in 1D. Understanding this principle is foundational for electromagnetism.