To solve this problem we will apply the concepts related to Coulomb's Law, for which electromagnetic force is defined as
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
k = Coulomb's Constant
[tex]q_{1,2}[/tex]= Magnitudes of the charges
r = Distance between the charges
From this equation and considering that the charges are constant, then we would have that the Force is inversely proportional to the distance of the bodies, that is,
[tex]\therefore F \propto \frac{1}{r^2}[/tex]
If we then make the comparison between two Forces that proportion would become
[tex]\frac{f_1}{f_2} = \frac{r_2^2}{r_1^2}[/tex]
We are looking for make this force twice as strong[tex](f_2 = 2f_1)[/tex], then
[tex]\frac{f_1}{2f_1} = \frac{r_2^2}{r_1^2}[/tex]
[tex]\frac{1}{2} = \frac{r_2^2}{r_1^2}[/tex]
[tex]\frac{1}{2} = (\frac{r_2}{r_1})^2[/tex]
[tex]\sqrt{\frac{1}{2}} = (\frac{r_2}{r_1})[/tex]
Rearranging to find the final distance
[tex]r_2 = \frac{r_1}{\sqrt{2}}[/tex]
As r = d,
[tex]d_2 = \frac{d}{\sqrt{2}}[/tex]
Therefore the distance would have to be changed to [tex]d_2 = \frac{d}{\sqrt{2}}[/tex]
In order to make the force twice as strong, the distance would have to be changed to d/√2
The correct answer to the question is Option B. d/√2
•Initial force (F₁) = F
•Initial distance (r₁) = d
•Final force (F₂) = 2F
•Final distance (r₂) =?
From Coulomb's law,
F = Kq₁q₂ / r²
If Kq₁q₂ is constant, then
F₁r₁² = F₂r₂²
With the above formula, we can obtain the final distance, r₂ as follow:
F₁r₁² = F₂r₂²
Fd² = 2Fr₂²
Divide both side by 2F
r₂² = Fd² / 2F
r₂² = d² / 2
Take the square root of both side
r₂= √(d² / 2)
r₂= d/√2
The correct answer to the question is Option B. d/√2
Complete question
When two point charges are a distance d apart, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed to
A) √2d
B) d/√2
C) d/4
D) 2d
E) d/2
Learn more about Coulomb's law:
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