A charge of `-3.00 mu C` is fixed at the center of a compass. Two additional charges are fixed on the circle of the compass (radius 0.100 m). The charges on the circle are `-4.00 mu C` at the position due north and `+5.00 mu C` at the position due east. what is the magnitude and direction of the net electrostatic force acting on the charge at the center ? Specify the direction relative to due east.

Options

A. `3.60 N, 51.3^(@)` north of east
B. `34.6 N, 36.3^(@)` north of east
C. `14.4 N, 44.2^(@)` north of east
D. `17.3 N, 38.7^(@)` south of east

Answer

To solve the problem, we need to calculate the net electrostatic force acting on the charge at the center due to the two charges fixed on the circle.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions:**
- Charge at the center (Q1): \( -3.00 \, \mu C = -3.00 \times 10^{-6} \, C \)
- Charge due North (Q2): \( -4.00 \, \mu C = -4.00 \times 10^{-6} \, C \)
- Charge due East (Q3): \( +5.00 \, \mu C = +5.00 \times 10^{-6} \, C \)
- Radius of the compass (r): \( 0.100 \, m \)

2. **Calculate the Forces Due to Each Charge:**
- **Force due to Q2 (F1):**
\[ F_1 = k \frac{|Q_1 \cdot Q_2|}{r^2} \]
Where \( k = 8.99 \times 10^9 \, N \cdot m^2/C^2 \).
\[ F_1 = 8.99 \times 10^9 \frac{|-3.00 \times 10^{-6} \cdot -4.00 \times 10^{-6}|}{(0.100)^2} \]
\[ F_1 = 8.99 \times 10^9 \frac{12.00 \times 10^{-12}}{0.01} = 8.99 \times 10^9 \cdot 1.2 \times 10^{-9} = 10.788 \, N \]
This force acts towards the south (since both charges are negative).

- **Force due to Q3 (F2):**
\[ F_2 = k \frac{|Q_1 \cdot Q_3|}{r^2} \]
\[ F_2 = 8.99 \times 10^9 \frac{|-3.00 \times 10^{-6} \cdot 5.00 \times 10^{-6}|}{(0.100)^2} \]
\[ F_2 = 8.99 \times 10^9 \frac{15.00 \times 10^{-12}}{0.01} = 8.99 \times 10^9 \cdot 1.5 \times 10^{-9} = 13.485 \, N \]
This force acts towards the east (since the center charge is negative and the east charge is positive).

3. **Determine the Net Force:**
- The forces \( F_1 \) and \( F_2 \) are perpendicular to each other (south and east).
- We can use the Pythagorean theorem to find the resultant force \( F_R \):
\[ F_R = \sqrt{F_1^2 + F_2^2} \]
\[ F_R = \sqrt{(10.788)^2 + (13.485)^2} = \sqrt{116.3 + 181.5} = \sqrt{297.8} \approx 17.26 \, N \]

4. **Calculate the Direction of the Resultant Force:**
- The angle \( \theta \) relative to the east can be found using:
\[ \tan(\theta) = \frac{F_1}{F_2} \]
\[ \tan(\theta) = \frac{10.788}{13.485} \]
\[ \theta = \tan^{-1}(0.799) \approx 38.66^\circ \]
- The direction of the resultant force is \( 38.66^\circ \) south of east.

### Final Answer:
The magnitude of the net electrostatic force acting on the charge at the center is approximately \( 17.26 \, N \) and the direction is \( 38.66^\circ \) south of east.