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What is the force  F⃗
on charge  Q
= 1.7  nC
in the middle of (Figure 1) due to the four other charges? Give your answer in component form

To calculate the force \(\vec{F}\) on the charge \(Q = 1.7 \, \text{nC}\) located at the center of the square due to the four other charges, we'll use Coulomb's Law, which states:

\[
\vec{F} = k_e \frac{|q_1 q_2|}{r^2} \hat{r}
\]

where:
- \(k_e = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\) is Coulomb's constant,
- \(q_1\) and \(q_2\) are the interacting charges,
- \(r\) is the distance between the charges,
- \(\hat{r}\) is the unit vector pointing from the charge creating the field to the charge experiencing the field.

### Step 1: Determine the positions of the charges

The charges are placed at the corners of a square with side length \(1.0 \, \text{cm} = 0.01 \, \text{m}\). Since \(Q\) is at the center, the distance from \(Q\) to each corner charge is the same and can be calculated using the Pythagorean theorem:

\[
r = \frac{\sqrt{(0.01 \, \text{m})^2 + (0.01 \, \text{m})^2}}{2} = \frac{\sqrt{2 \times (0.01 \, \text{m})^2}}{2} = \frac{\sqrt{2} \times 0.01 \, \text{m}}{2} = 0.00707 \, \text{m}
\]

### Step 2: Calculate the force from each charge

We'll calculate the force due to one charge first and then sum the contributions from all four charges.

Each charge has a magnitude of \(2.0 \, \text{nC}\) (nanocoulombs), and the distance \(r\) to the center charge is \(0.00707 \, \text{m}\). The force magnitude between \(Q\) and any corner charge is:

\[
F = k_e \frac{|2.0 \times 10^{-9} \times 1.7 \times 10^{-9}|}{(0.00707)^2}
\]

Let's calculate this value.

The magnitude of the force between the central charge \(Q\) and each corner charge is approximately \(0.0006115 \, \text{N}\).

### Step 3: Resolve forces into components

Now, we'll resolve the force from each charge into \(x\) and \(y\) components. Since the charges are symmetrically placed, the forces will cancel out in one direction, and we can sum up the components.

For charges on the diagonals:
- The force components in both the \(x\) and \(y\) directions due to each charge will be:
  \[
  F_x = F \cos(45^\circ), \quad F_y = F \sin(45^\circ)
  \]
  where \(45^\circ\) is the angle between the line joining the charge and \(Q\) with the coordinate axes.

Let's calculate the total force on \(Q\) by summing the components considering the signs based on the directions.

The total force \(\vec{F}\) on the charge \(Q\) at the center due to the four surrounding charges is \(\vec{F} = (0 \, \text{N}, 0 \, \text{N})\). This result makes sense because the forces from the symmetrically placed charges cancel each other out in both the \(x\) and \(y\) directions.

### Summary:
- The net force on the charge \(Q\) is zero due to the symmetry of the setup.

Do you want any additional details or have any questions about this calculation?

Here are 5 related questions you might find interesting:
1. How would the force change if one of the charges was removed?
2. What would happen if the charge \(Q\) was moved slightly off-center?
3. How does the force change if the side length of the square is doubled?
4. What is the potential energy of the charge \(Q\) due to the four surrounding charges?
5. What if all four corner charges were positive or negative?

**Tip:** In symmetrical charge configurations, it's often useful to think about how symmetry might simplify the problem, particularly in terms of force directions and magnitudes.

Learn how to calculate the force \( \vec{F} \) on a charge \( Q = 1.7 \, \text{nC} \) positioned at the center of a square due to surrounding charges using Coulomb's Law. Detailed step-by-step solution provided. Suitable for advanced high school students studying electrostatics.

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