Question

1. coulombs law states that the force between two stationary particles is determined by the magnitude of each charge and the distance between their centers. this equation can be written in the form ( f = k\frac{q_1q_2}{r^2} ), where ( f ) is the magnitude of the electrostatic force, ( q_1 ) and ( q_2 ) are the charges of the two particles, ( k ) is an electrostatic constant and ( r ) is the distance between the two centers of the particles. write an equation for ( r ) in terms of ( f, k, q_1 ) and ( q_2 ). 10. coulombs law states that the force between two stationary particles is determined by the magnitude of each charge and the distance between their centers. this equation can be written in the form ( f = k\frac{q_1q_2}{r^2} ), where ( f ) is the magnitude of the electrostatic force, ( q_1 ) and ( q_2 ) are the charges of the two particles, ( k ) is an electrostatic constant and ( r ) is the distance between the two centers of the particles. describe the result on the force if the distance between the particles doubles, the charge of the first particle quadruples and the charge of the second particle decreases by 33%.

Explanation:

Question 9

Step1: Start with Coulomb's Law formula

We have the formula \( F = k\frac{Q_1Q_2}{r^2} \). Our goal is to solve for \( r \) in terms of \( F \), \( k \), \( Q_1 \), and \( Q_2 \). First, we can multiply both sides of the equation by \( r^2 \) to get rid of the denominator on the right - hand side.
\( F\times r^2=k\frac{Q_1Q_2}{r^2}\times r^2 \)
Simplifying the right - hand side, we have \( F r^2=kQ_1Q_2 \)

Step2: Isolate \( r^2 \)

Next, we divide both sides of the equation \( F r^2 = kQ_1Q_2 \) by \( F \) to isolate \( r^2 \).
\( r^2=\frac{kQ_1Q_2}{F} \)

Step3: Solve for \( r \)

To solve for \( r \), we take the square root of both sides of the equation \( r^2=\frac{kQ_1Q_2}{F} \). Since \( r \) represents a distance, we consider the positive square root (as distance cannot be negative in this context).
\( r = \sqrt{\frac{kQ_1Q_2}{F}} \) or we can also write it as \( r=\frac{\sqrt{kQ_1Q_2}}{\sqrt{F}} \) (rationalizing the denominator is not necessary here, but both forms are correct)

Let the original force be \( F = k\frac{Q_1Q_2}{r^2} \)

Step1: Define the new values

• The distance between the particles doubles, so the new distance \( r'=2r \)
• The charge of the first particle quadruples, so the new charge \( Q_1' = 4Q_1 \)
• The charge of the second particle decreases by 33%. A 33% decrease means the new charge is \( (1 - 0.33)Q_2=0.67Q_2\approx\frac{2}{3}Q_2 \) (we can also use the exact value of \( 0.67 \) or \( \frac{2}{3} \) for calculation. Let's use the exact percentage calculation: 33% is approximately \( \frac{1}{3} \), so the remaining charge is \( Q_2'=(1-\frac{1}{3})Q_2=\frac{2}{3}Q_2 \))

Step2: Substitute the new values into the formula

The new force \( F'=k\frac{Q_1'Q_2'}{(r')^2} \)
Substitute \( Q_1' = 4Q_1 \), \( Q_2'=\frac{2}{3}Q_2 \) and \( r' = 2r \) into the formula:
\( F'=k\frac{(4Q_1)\times(\frac{2}{3}Q_2)}{(2r)^2} \)

Step3: Simplify the expression

First, simplify the numerator and the denominator:

• The numerator: \( (4Q_1)\times(\frac{2}{3}Q_2)=\frac{8}{3}Q_1Q_2 \)
• The denominator: \( (2r)^2 = 4r^2 \)

So \( F'=k\frac{\frac{8}{3}Q_1Q_2}{4r^2} \)
We can rewrite this as \( F'=k\times\frac{8}{3}Q_1Q_2\times\frac{1}{4r^2} \)
Simplify the constants: \( \frac{8}{3}\times\frac{1}{4}=\frac{2}{3} \)
So \( F'=\frac{2}{3}k\frac{Q_1Q_2}{r^2} \)

But we know that the original force \( F = k\frac{Q_1Q_2}{r^2} \), so we can substitute \( F \) into the equation for \( F' \):
\( F'=\frac{2}{3}F \)

This means that the new force is \( \frac{2}{3} \) of the original force, or we can say that the force decreases by \( 1-\frac{2}{3}=\frac{1}{3}\approx33.3\% \)

Answer:

\( r=\sqrt{\frac{kQ_1Q_2}{F}} \)

Question 10