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Chapter 9: Problem 7

Two identical point charges \(q_1\) and \(q_2\) are at a distance \(r\) apart. If the size of \(q_1\) is doubled and the distance between them tripled, the strength of the electrical force between them (A) goes up by a factor of 3 . (B) goes down by a factor of 3 . (C) goes down by a factor of 9 . (D) goes down by a factor of \(2 / 3\). (E) goes down by a factor of \(2 / 9\)

Short Answer

Expert verified
Question: After doubling the size of one of the point charges and tripling the distance between them, the electrical force between the two point charges will: (A) remain the same (B) go up by a factor of 2 (C) go up by a factor of 6 (D) go down by a factor of 6 (E) go down by a factor of 2/9

Step by step solution

01

Understanding Coulomb's Law

Coulomb's law states that the electrical force (\(F\)) between two point charges \(q_1\) and \(q_2\) is proportional to the product of their charges and inversely proportional to the square of the distance (\(r\)) between them. Mathematically, this is expressed as: $$ F = k \frac{q_1q_2}{r^2} $$ Where \(k\) is Coulomb's constant.
02

Calculate the initial electrical force

We can use Coulomb's law to calculate the initial electrical force between \(q_1\) and \(q_2\): $$ F_1 = k \frac{q_1q_2}{r^2} $$
03

Calculate the electrical force after the changes

Now, let's double the size of \(q_1\) (i.e., \(2q_1\)) and triple the distance between them (i.e., \(3r\)). We will now calculate the new electrical force \(F_2\) using these changes: $$ F_2 = k \frac{(2q_1)q_2}{(3r)^2} $$
04

Simplify the new electrical force equation

We can simplify the fraction in the equation: $$ F_2 = k \frac{2q_1q_2}{9r^2} $$
05

Compare initial and new electrical force

Notice that the initial electrical force equation was: $$ F_1 = k \frac{q_1q_2}{r^2} $$ Comparing the two equations, we can see the relationship between the initial electrical force \(F_1\) and the new electrical force \(F_2\): $$ F_2 = \frac{2}{9} F_1 $$ This means that the new electrical force \(F_2\) is reduced by a factor of \(2/9\) as compared to the initial electrical force \(F_1\). So, the correct answer is: (E) goes down by a factor of \(2/9\).

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Most popular questions from this chapter

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