When two point charges are 2.0 cm apart, each one experiences a 1.0-Nelectric force due to the other charge. If they are moved to a new separation of 8.0 cm,the electric force on each of them is closest to

The electric force between two point charges is given by Coulomb's Law:

\[
F = k \frac{|q_1 q_2|}{r^2}
\]

Where:
- \(F\) is the electric force between the charges,
- \(k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2\) is Coulomb's constant,
- \(q_1\) and \(q_2\) are the magnitudes of the two point charges,
- \(r\) is the distance between the charges.

### Step 1: Determine the relationship between force and distance.
From Coulomb's law, we can see that the electric force \(F\) is inversely proportional to the square of the distance \(r\) between the charges:

\[
F \propto \frac{1}{r^2}
\]

This means that if the distance between the charges is increased, the force decreases by the square of the ratio of the distances.

### Step 2: Apply the initial and new conditions.
- Initially, the distance between the charges is 2.0 cm (0.02 m) and the force is 1.0 N.
- The new distance between the charges is 8.0 cm (0.08 m).

We want to find the new force \(F_{\text{new}}\) when the separation is changed. Using the inverse square law:

\[
\frac{F_{\text{new}}}{F_{\text{initial}}} = \left(\frac{r_{\text{initial}}}{r_{\text{new}}}\right)^2
\]

Substitute the given values:

\[
\frac{F_{\text{new}}}{1.0 \, \text{N}} = \left(\frac{0.02 \, \text{m}}{0.08 \, \text{m}}\right)^2
\]

\[
F_{\text{new}} = 1.0 \, \text{N} \times \left(\frac{0.02}{0.08}\right)^2
\]

\[
F_{\text{new}} = 1.0 \times \left(\frac{1}{4}\right)^2
\]

\[
F_{\text{new}} = 1.0 \times \frac{1}{16} = 0.0625 \, \text{N}
\]

### Final Answer:
The new electric force on each charge, when the separation is increased to 8.0 cm, is **0.0625 N**.

Would you like further clarification or additional details on the problem?

### Related Questions:
1. How does Coulomb’s Law apply to the force between three or more point charges?
2. How can you calculate the electric potential energy between two point charges?
3. What would happen if one of the charges were doubled in magnitude?
4. How do electric fields relate to electric forces between point charges?
5. How does the medium between charges affect the force?

**Tip:** Remember, when dealing with electric forces, always check if the charges are in a vacuum or another medium, as this will affect the constant \(k\) in Coulomb’s Law.

When two point charges are 2.0 cm apart, each one experiences a 1.0-N electric force due to the other charge. If they are moved to a new separation of 8.0 cm, the electric force on each of them is closest to

Learn how to calculate the electric force between two point charges using Coulomb's Law. This example shows how the force changes when the separation between charges increases from 2 cm to 8 cm, resulting in a force of 0.0625 N. Perfect for students in Grades 10-12 studying electrostatics.

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