Change in Electric Force due to Distance Variation

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Two charges $q_1 = 7,2Q$ and $q_2 = 8,8Q$ are separated by a distance $d$. Assume that the electric force between them is $F$. If the separation between two charges is changed to $0,87d$, what would be the new electric force between them in terms of $F$. Round your answer to two decimal places.

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1
Step 1

In this physics problem, we need to find how the electric force between two charges changes when the distance between them is modified. Let's start by listing our given variables.

Electrostatic Force Analysis

2
Step 2

Initially, we have two charges, queue one and queue two, separated by a distance d. The resulting electric force is given as F.

Initial State

$$q_1 = 7.2Q, \quad q_2 = 8.8Q$$
$$r_1 = d$$
$$F_1 = F$$
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Step 3

According to Coulomb's Law, the force is proportional to the product of the charges and inversely proportional to the square of the distance between them.

$$F = k \frac{|q_1 q_2|}{r^2}$$
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Step 4

Now, let's look at the final state. The separation distance is changed to zero point eight seven times the original distance d.

Final State

$$q_1' = 7.2Q, \quad q_2' = 8.8Q$$
$$r_2 = 0.87d$$
$$F_2 = ?$$
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Step 5

Notice that the charges remain exactly the same. Only the distance changes. This allows us to use a ratio approach between the two forces.

$$F \propto \frac{1}{r^2}$$
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Step 6

We can write the relationship as the new force F sub two over the old force F sub one equals the inverse square of their distances.

$$\frac{F_2}{F_1} = \left( \frac{r_1}{r_2} \right)^2$$
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Step 7

Substituting our values, F sub one is F, the initial distance r sub one is d, and the final distance r sub two is zero point eight seven d.

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About This Question

Subject
Physics
Topic
Electrostatics
Difficulty
Medium
Exam
STEM
Question Type
Open Ended

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