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Anna007 [38]
3 years ago
8

Suppose you wanted to increase the force between two point charges by a factor of 5. By what factor must you change the distance

between them
Physics
1 answer:
zhannawk [14.2K]3 years ago
7 0

Answer:

the distance between the two point charges will be reduced by a factor of  \frac{1}{\sqrt{5} } = 0.447

Explanation:

The force between two point charges is given by Coulomb's law;

F = \frac{kQ_1Q_2}{r^2}

where;

k is Coulomb's constant

Q₁ and Q₂ are two point charges

r is the distance between the two charges

From the equation above, the following relationship between force and distance can be deduced;

F₁r₁² = F₂r₂²

F₁r₁² = (5F₁)r₂²

r₁² = 5r₂²

r_2^2 = \frac{r_1^2}{5} \\\\r_2 = \sqrt{\frac{r_1^2}{5}} \\\\r_2 = \frac{r_1}{\sqrt{5} } \\\\r_2 = \frac{1}{\sqrt{5} } \ r_1\\\\r_2 = 0.447( r_1)

Thus, the distance between the two point charges will be reduced by a factor of  \frac{1}{\sqrt{5} } = 0.447

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Answer:

Fa=774 N

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Explanation:

We will solve this problem by equating forces on each axis.

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⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

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⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

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⇒Fa=2.24*Fb

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3 years ago
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4 0
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