In this problem, we will use vector notation as (x,y,z). A charge q1 = 5.2 μC is located at (5.9 cm, 13.2 cm, 10.4 cm), and another charge q2 = 4.2 μC is located at (22.5 cm, 2.8 cm, 3.1 cm). What is the electric field at the origin?

Enet = ( N/C,  N/C,  N/C).


To calculate the electric field \( \mathbf{E}_{\text{net}} \) at the origin due to two point charges, we can use the following formula for the electric field due to a point charge:

\[
\mathbf{E} = \frac{k_e \cdot q}{r^2} \cdot \hat{r}
\]

where:
- \( k_e \) is Coulomb's constant, \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \),
- \( q \) is the charge (in Coulombs),
- \( r \) is the distance from the charge to the point where the field is being calculated (in meters),
- \( \hat{r} \) is the unit vector from the charge to the point.

### Step 1: Convert units
Convert all distances to meters since Coulomb's constant is in SI units.

- Charge \( q_1 = 5.2 \, \mu C = 5.2 \times 10^{-6} \, C \)
- Charge \( q_2 = 4.2 \, \mu C = 4.2 \times 10^{-6} \, C \)
- Position of \( q_1 \) is \( (5.9 \, \text{cm}, 13.2 \, \text{cm}, 10.4 \, \text{cm}) = (0.059 \, \text{m}, 0.132 \, \text{m}, 0.104 \, \text{m}) \)
- Position of \( q_2 \) is \( (22.5 \, \text{cm}, 2.8 \, \text{cm}, 3.1 \, \text{cm}) = (0.225 \, \text{m}, 0.028 \, \text{m}, 0.031 \, \text{m}) \)

### Step 2: Calculate distance and unit vectors
For each charge, calculate the distance from the origin and the unit vector pointing from the charge to the origin.

#### For charge \( q_1 \):
- The distance \( r_1 \) from the origin is:

\[
r_1 = \sqrt{(0.059)^2 + (0.132)^2 + (0.104)^2}
\]

- The unit vector \( \hat{r}_1 \) pointing from \( q_1 \) to the origin is:

\[
\hat{r}_1 = \frac{-\mathbf{r}_1}{r_1} = \left( \frac{-0.059}{r_1}, \frac{-0.132}{r_1}, \frac{-0.104}{r_1} \right)
\]

#### For charge \( q_2 \):
- The distance \( r_2 \) from the origin is:

\[
r_2 = \sqrt{(0.225)^2 + (0.028)^2 + (0.031)^2}
\]

- The unit vector \( \hat{r}_2 \) pointing from \( q_2 \) to the origin is:

\[
\hat{r}_2 = \frac{-\mathbf{r}_2}{r_2} = \left( \frac{-0.225}{r_2}, \frac{-0.028}{r_2}, \frac{-0.031}{r_2} \right)
\]

### Step 3: Calculate electric field components
For each charge, use the formula \( \mathbf{E} = \frac{k_e \cdot q}{r^2} \cdot \hat{r} \) to calculate the electric field at the origin.

Let me calculate the values.It seems like I can’t do more advanced data analysis right now. Please try again later. However, if you'd like, I can walk you through the manual process for solving this or help you with any other questions.

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Here are 5 related questions to explore further:
1. How would the electric field change if one of the charges were negative?
2. What is the relationship between electric field strength and distance from a point charge?
3. How would the result differ if the point of interest was not at the origin?
4. What are the units of the electric field, and how do they relate to other physical quantities?
5. How does the superposition principle apply to electric fields from multiple charges?

**Tip:** When calculating the electric field from multiple charges, always resolve the fields into components and sum them vectorially for accuracy.

In this problem, we will use vector notation as (x,y,z). A charge q1 = 5.2 μC is located at (5.9 cm, 13.2 cm, 10.4 cm), and another charge q2 = 4.2 μC is located at (22.5 cm, 2.8 cm, 3.1 cm). What is the electric field at the origin?

Explore how to calculate the electric field at the origin due to two point charges using Coulomb's Law. This detailed guide covers vector notation, electric field equations, and superposition, perfect for undergraduate physics students.

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