A2 kg mass having a charge of +5 µC and a 1 kg mass with charge-2 uC are separated by 0.5 m as shown below. The two masses are initially held in place and then released. What is the magnitude of the acceleration of the 2 kg mass the moment after they are released? Neglect the effects of gravity. Suc, 2 kg O0.18 m/s² 0.36 m/s² O1.42 m/s² O2.45 m/s² 09.8 m/s² -2 μC, 1 kg
A2 kg mass having a charge of +5 µC and a 1 kg mass with charge-2 uC are separated by 0.5 m as shown below. The two masses are initially held in place and then released. What is the magnitude of the acceleration of the 2 kg mass the moment after they are released? Neglect the effects of gravity. Suc, 2 kg O0.18 m/s² 0.36 m/s² O1.42 m/s² O2.45 m/s² 09.8 m/s² -2 μC, 1 kg
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![### Physics Problem - Coulomb's Law Application
#### Question 14 of 22
A 2 kg mass having a charge of +5 μC and a 1 kg mass with charge -2 μC are separated by 0.5 m as shown below. The two masses are initially held in place and then released. What is the magnitude of the acceleration of the 2 kg mass the moment after they are released? Neglect the effects of gravity.
![Diagram]
```
+5 μC, 2 kg -2 μC, 1 kg
* ----------- 0.5 m ------------ *
```
#### Multiple Choice Options
- O (0.18 m/s²)
- O (0.36 m/s²)
- O (1.42 m/s²)
- O (2.45 m/s²)
- O (9.8 m/s²)
##### Explanation:
To solve this problem, we'll apply Coulomb's Law, which states that the magnitude of the electrostatic force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:
\[ F = \frac{k |q_1 q_2|}{r^2} \]
Where:
- \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \))
- \( q_1 = +5 \, \mu\text{C} = +5 \times 10^{-6} \, \text{C} \)
- \( q_2 = -2 \, \mu\text{C} = -2 \times 10^{-6} \, \text{C} \)
- \( r = 0.5 \, \text{m} \)
Calculate the electrostatic force \( F \):
\[ F = \frac{(8.99 \times 10^9) \times (5 \times 10^{-6}) \times (2 \times 10^{-6})}{(0.5)^2} \]
\[ F = \frac{8.99 \times 10^9 \times 10 \times 10^{-12}}{0.25} \]
\[ F = \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F45c459d4-c7f0-416a-8b1d-00b29b90700e%2F5dcefdb1-a3dd-4c01-840f-67f68289e4ea%2F0anqqw9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics Problem - Coulomb's Law Application
#### Question 14 of 22
A 2 kg mass having a charge of +5 μC and a 1 kg mass with charge -2 μC are separated by 0.5 m as shown below. The two masses are initially held in place and then released. What is the magnitude of the acceleration of the 2 kg mass the moment after they are released? Neglect the effects of gravity.
![Diagram]
```
+5 μC, 2 kg -2 μC, 1 kg
* ----------- 0.5 m ------------ *
```
#### Multiple Choice Options
- O (0.18 m/s²)
- O (0.36 m/s²)
- O (1.42 m/s²)
- O (2.45 m/s²)
- O (9.8 m/s²)
##### Explanation:
To solve this problem, we'll apply Coulomb's Law, which states that the magnitude of the electrostatic force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:
\[ F = \frac{k |q_1 q_2|}{r^2} \]
Where:
- \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \))
- \( q_1 = +5 \, \mu\text{C} = +5 \times 10^{-6} \, \text{C} \)
- \( q_2 = -2 \, \mu\text{C} = -2 \times 10^{-6} \, \text{C} \)
- \( r = 0.5 \, \text{m} \)
Calculate the electrostatic force \( F \):
\[ F = \frac{(8.99 \times 10^9) \times (5 \times 10^{-6}) \times (2 \times 10^{-6})}{(0.5)^2} \]
\[ F = \frac{8.99 \times 10^9 \times 10 \times 10^{-12}}{0.25} \]
\[ F = \frac
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