To evaluate how the electrical force between two charges changes when their magnitudes are altered, we can use Coulomb's Law, which states:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where \(F\) is the electrical force between two charges, \(k\) is Coulomb's constant, \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between the charges.
Let the original charges be \(q_1\) and \(q_2\). The original force \(F_{\text{original}}\) can be expressed as:
\[ F_{\text{original}} = k \frac{|q_1 q_2|}{r^2} \]
According to the problem, the first charge is decreased to one-third of its original value, and the second charge is decreased to one-half of its original value. Therefore, the new charges are:
- New \(q_1' = \frac{1}{3} q_1\)
- New \(q_2' = \frac{1}{2} q_2\)
Now we can calculate the new force \(F_{\text{new}}\):
\[ F_{\text{new}} = k \frac{|q_1' q_2'|}{r^2} = k \frac{\left|\frac{1}{3} q_1 \cdot \frac{1}{2} q_2\right|}{r^2} \]
This simplifies to:
\[ F_{\text{new}} = k \frac{|q_1 q_2|}{r^2} \cdot \frac{1}{3 \cdot 2} = F_{\text{original}} \cdot \frac{1}{6} \]
This shows that the new force \(F_{\text{new}}\) is one-sixth of the original force.
Thus, the electrical force between the charges will decrease to one-sixth the original force.
The correct answer is: It will decrease to one-sixth the original force.
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