Respuesta :
Answer:
2F
Explanation:
From the Coulomb's Law, we know that the force between two point charges separated by a distance r is
[tex]F=k\frac{q_1q_2}{r^2}[/tex]
As the magnitude of the charges and distance both changed in the second part of the question, the new force will be
[tex]F_2=k\frac{2q_1\times 4q_2}{(2r)^2}[/tex]
[tex]F_2=2k\frac{q_1q_2}{r^2}[/tex]
Thus upon changing the charges and distance between the charges the magnitude of the new force will be two times the magnitude of earlier force, i.e. 2F.
The new magnitude of electrostatic force between the charges 2q1 and 2q2 is F. Hence, option (1) is correct.
Electrostatic Force:
The force offered by the charged entities, on each other is known as the electrostatic force.
Given data:
The magnitude of two point charges is, q1 and q2.
The distance of separation between the charges is, r.
The magnitude of electrostatic force between the charges is, F.
The mathematical expression for the electrostatic force is given as,
[tex]F = \dfrac{k \times q1 \times q2}{r^{2}}[/tex] ...............................................................(1)
Here, k is the Coulomb's constant.
Now, if the magnitude of point charges are 2q1 and 2q2 respectively, such that the distance between them is 2r. Then, the electrostatic force becomes,
[tex]F' = \dfrac{k \times 2q1 \times 2q2}{(2r)^{2}}\\\\F' = 4 \times \dfrac{k \times q1 \times q2}{4r^{2}}\\\\F' = \dfrac{k \times q1 \times q2}{r^{2}}[/tex]
Substituting the value of equation (1) to above expression to get,
F' = F
Thus, we can conclude that the new magnitude of electrostatic force between the charges 2q1 and 2q2 is F. Hence, option (1) is correct.
Learn more about the electrostatic force here:
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