Coulomb's law states that the electric force \( F \) between two point charges can be calculated using the formula:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where:
- \( k = 8.99 \times 10^9 , \text{N m}^2/\text{C}^2 \) (Coulomb's constant),
- \( q_1 \) and \( q_2 \) are the charges,
- \( r \) is the distance between the charges.
In this case, we have two negative charges, both equal to \( e = 1.6 \times 10^{-19} , \text{C} \). Since both charges are negative, the force will be repulsive, hence it will be negative.
Let's calculate the force:
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Use the values: \[ q_1 = -e = -1.6 \times 10^{-19} , \text{C} \] \[ q_2 = -e = -1.6 \times 10^{-19} , \text{C} \] \[ r = 10^{-14} , \text{m} \]
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Insert values into Coulomb's law: \[ F = k \frac{|q_1 q_2|}{r^2} = 8.99 \times 10^9 \frac{(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{(10^{-14})^2} \]
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Calculate: \[ F = 8.99 \times 10^9 \frac{(2.56 \times 10^{-38})}{(1 \times 10^{-28})} \] \[ F = 8.99 \times 10^9 \times 2.56 \times 10^{-10} \] \[ F = 2.3 , \text{N} \]
Since both charges are negative, the force is repulsive, which means we express it as a negative value:
\[ F = -2.3 , \text{N} \]
Thus, the electrical force between these two particles is −2.3 N.
2 answers