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Chapter 21: Problem 34

Two charged spheres are \(8 \mathrm{~cm}\) apart. They are moved closer to each other enough that the force on each of them increases four times. How far apart are they now?

Short Answer

Expert verified
Answer: The new distance between the charged spheres is 4 cm.

Step by step solution

01

Write down the initial conditions and the variables to find

The initial distance between the charged spheres is \(8 \mathrm{~cm}\). We are looking for the new distance after the force increases four times. Let's denote the initial distance as \(d_1\) and the final distance as \(d_2\). We have: - \(d_1 = 8 \mathrm{~cm}\) - Find \(d_2\)
02

Write down the Coulomb's Law formula

The formula for the force between two charged particles according to Coulomb's Law is: $$F = k \frac{q_1q_2}{d^2}$$ Where: - \(F\) is the force between the two charged particles. - \(k\) is the Coulomb's constant (\(8.99 \times 10^9 \mathrm{Nm^2C^{-2}}\)) - \(q_1\) and \(q_2\) are the charges of the particles. - \(d\) is the distance between the charged particles.
03

Use the ratio of the forces given in the problem

We are given that the force increases by four times after the spheres are moved closer. Let's set up a ratio to find the relationship between the initial and final distances: $$\frac{F_1}{F_2} = \frac{4}{1}$$ Since these forces depend on the distance between the spheres according to Coulomb's Law, we can plug in the formula for each force: $$\frac{k\frac{q_1q_2}{d_1^2}}{k\frac{q_1q_2}{d_2^2}} = \frac{4}{1}$$
04

Simplify the equation and solve for \(d_2\)

Notice that \(k\) and \(q_1q_2\) are present in both the numerator and the denominator, so we can cancel them out to simplify the equation: $$\frac{1}{d_1^2} \div \frac{1}{d_2^2} = 4$$ Now cross-multiply and solve for \(d_2\): $$4d_2^2 = d_1^2$$ We know \(d_1 = 8 \mathrm{~cm}\), so we can substitute that into our equation and solve for \(d_2\): $$4d_2^2 = (8 \mathrm{~cm})^2$$ $$d_2^2 = \frac{64 \mathrm{~cm}^2}{4}$$ $$d_2^2 = 16 \mathrm{~cm}^2$$ Now, take the square root of both sides to find \(d_2\): $$d_2 = \sqrt{16 \mathrm{~cm}^2}$$ $$d_2 = 4 \mathrm{~cm}$$
05

State the final answer

After moving the charged spheres closer to each other such that the force on each increases four times, they are now \(4 \mathrm{~cm}\) apart.

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Most popular questions from this chapter

Four point charges are placed at the following \(x y\) coordinates: \(Q_{1}=-1 \mathrm{mC},\) at \((-3 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{2}=-1 \mathrm{mC},\) at \((+3 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{3}=+1.024 \mathrm{mC},\) at \((0 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{4}=+2 \mathrm{mC},\) at \((0 \mathrm{~cm},-4 \mathrm{~cm})\) Calculate the net force on charge \(Q_{4}\) due to charges \(Q_{1}, Q_{2}\) and \(Q_{3}\).

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