the electric field produced by a positively charged particle, placed in an xy-plane is `7.2 (4i+3j) N//C` at the point (3 cm, 3 cm) and `100 hat(i) N//C` at the point (2 cm, 0)

Options

A. The x-coordinate of the charged particle is -2cm.
B. The charged particle is placed on the x-axis
C. The charge of the particle is `10xx10^(-12) C`.
D. The electric potential at the origin due to the charge is 9V.

Answer

To solve the problem, we need to determine the position of the charged particle and its magnitude based on the electric fields given at specific points in the xy-plane.

### Step-by-Step Solution:

1. **Understanding Electric Field Direction**:
The electric field produced by a positively charged particle points radially outward from the charge. Given the electric field at point (2 cm, 0) is `100 i N/C`, it indicates that the charge is located to the left of this point along the x-axis.

2. **Identifying the Charge's Position**:
Let’s denote the position of the charge as \( (x_0, 0) \). The distance from the charge to the point (2 cm, 0) is \( 2 - x_0 \). The electric field \( E \) at this point can be expressed using Coulomb's law:
\[ E = \frac{k \cdot Q}{(2 - x_0)^2} \]
where \( k \) is Coulomb's constant and \( Q \) is the charge.

3. **Substituting Known Values**:
We know that \( E = 100 \, \text{N/C} \). Thus, we can write:
\[ 100 = \frac{9 \times 10^9 \cdot Q}{(2 - x_0)^2} \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ (2 - x_0)^2 = \frac{9 \times 10^9 \cdot Q}{100} \]
\[ (2 - x_0)^2 = 9 \times 10^7 \cdot Q \]

5. **Analyzing the Electric Field at (3 cm, 3 cm)**:
The electric field at point (3 cm, 3 cm) is given as \( 7.2 (4i + 3j) \, \text{N/C} \). We can break this down into components:
\[ E_x = 7.2 \cdot 4 = 28.8 \, \text{N/C}, \quad E_y = 7.2 \cdot 3 = 21.6 \, \text{N/C} \]

6. **Finding the Angle**:
The angle \( \theta \) made with the x-axis can be found using:
\[ \tan \theta = \frac{E_y}{E_x} = \frac{21.6}{28.8} = \frac{3}{4} \]
Thus, \( \theta = \tan^{-1}\left(\frac{3}{4}\right) \).

7. **Using the Relation Between Electric Field and Distance**:
The electric field at (3 cm, 3 cm) can also be expressed as:
\[ E = \frac{k \cdot Q}{r^2} \]
where \( r \) is the distance from the charge to the point (3 cm, 3 cm). The distance \( r \) can be calculated using the distance formula:
\[ r = \sqrt{(3 - x_0)^2 + (3 - 0)^2} \]

8. **Setting Up the Equation**:
We can set up the equation for the electric field at this point:
\[ 7.2(4i + 3j) = \frac{k \cdot Q}{((3 - x_0)^2 + 3^2)} \]

9. **Solving for Charge \( Q \)**:
Using both equations derived from the electric fields at (2 cm, 0) and (3 cm, 3 cm), we can solve for \( Q \) and \( x_0 \).

10. **Calculating Electric Potential at Origin**:
Finally, the electric potential \( V \) at the origin due to the charge \( Q \) can be calculated using:
\[ V = \frac{k \cdot Q}{r} \]
where \( r \) is the distance from the charge to the origin.