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Chapter 11: Problem 1

If the distance between two positive point charges is tripled, then the strength of the electrostatic repulsion between them will decrease by a factor of (A) 3 (B) 6 (C) 8 (D) 9

Short Answer

Expert verified
The strength of the electrostatic repulsion between the two charges decreases by a factor of 9. Hence, the correct answer is (D) 9.

Step by step solution

01

Understanding the Concept

From Coulomb's law, the force (F) between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The formula for force is given by: F = k * (Q1 * Q2 / r^2), where Q1 and Q2 are the magnitudes of the charges, r is the distance between them, and k is a constant.
02

Applying the Change

According to the given problem, the distance between the charges is tripled. So, the new distance between them is 3r. Substituting this into the formula, the new force (F') is given by: F' = k * (Q1 * Q2 / (3r)^2).
03

Calculating the decrease factor

We now need to find by what factor the strength of the electrostatic force decreases. This is given by the ratio F / F', which simplifies to: (F / F') = (k * (Q1 * Q2 / r^2)) / (k * (Q1 * Q2 / (3r)^2)) = (3r)^2 / r^2 = 9. Therefore, the strength of the force decreases by a factor of 9.

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Most popular questions from this chapter

A charge of \(-3 Q\) is transferred to a solid metal sphere of radius \(r\). Where will this excess charge reside? (A) \(-Q\) at the center, and \(-2 Q\) on the outer surface (B) \(-3 Q\) at the center (C) \(-3 Q\) on the outer surface (D) \(-Q\) at the center, \(-Q\) in a ring of radius \(\frac{1}{2} r,\) and \(-Q\) on the outer surface

Two charges \(\left(q_{1}\right.\) and \(\left.q_{2}\right)\) are separated by a distance \(r .\) If the ratio of \(F_{\mathrm{G}} / F_{\mathrm{E}}\) is equal to \(9.0 \times 10^{43}\), what is the new ratio if the distance between the two charges is now \(3 r\) ? (A) \(1.0 \times 10^{43}\) (B) \(3.0 \times 10^{43}\) (C) \(9.0 \times 10^{43}\) (D) \(27.0 \times 10^{43}\)

An object of charge \(+q\) feels an electric force \(\mathbf{F}_{\mathrm{E}}\) when placed at a particular location in an electric field, E. Therefore, if an object of charge \(-2 q\) were placed at the same location where the first charge was, it would feel an electric force of (A) \(\frac{-\mathbf{F}_{\mathrm{E}}}{2}\) (B) \(-2 \mathbf{F}_{\mathrm{E}}\) (C) \(-2 q \mathbf{F}_{\mathrm{E}}\) (D) \(\frac{-2 \mathbf{F}_{\mathrm{E}}}{q}\)

How far apart are two charges \(\left(q_{1}=8 \times 10^{-6} \mathrm{C}\right.\) and \(q_{2}=6 \times\) \(10^{-6} \mathrm{C}\) ) if the electric force exerted by the charges on each other has a magnitude of \(2.7 \times 10^{-2} \mathrm{~N}\) ? (A) \(1 \mathrm{~m}\) (B) \(2 \mathrm{~m}\) (C) \(3 \mathrm{~m}\) (D) \(4 \mathrm{~m}\)

Two 1 kg spheres each carry a charge of magnitude 1 C. How does \(F_{\mathrm{E}}\), the strength of the electric force between the spheres, compare to \(F_{\mathrm{G}}\), the strength of their gravitational attraction? (A) \(F_{\mathrm{E}}F_{\mathrm{G}}\) (D) If the charges on the spheres are of the same sign, then \(F_{\mathrm{E}}\) \(>F_{\mathrm{G}}\); but if the charges on the spheres are of the opposite sign, then \(F_{\mathrm{E}}
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