To find the magnitude and direction of the electrical force applied by the negative charge \( q_1 \) on the positive charge \( q_2 \), we can use Coulomb's law, which is given by the formula:
\[ F_e = k \frac{|q_1 q_2|}{r^2} \]
where:
- \( F_e \) is the magnitude of the electrical force,
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 , \text{N m}^2/\text{C}^2 \),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.
Given:
- \( q_1 = -6 , \mu C = -6 \times 10^{-6} , C \),
- \( q_2 = 3 , \mu C = 3 \times 10^{-6} , C \),
- \( r = 0.002 , m \).
Now substituting the values into the formula:
\[ F_e = (8.99 \times 10^9) \frac{|(-6 \times 10^{-6})(3 \times 10^{-6})|}{(0.002)^2} \]
Calculate \( |q_1 q_2| \):
\[ |(-6 \times 10^{-6})(3 \times 10^{-6})| = 18 \times 10^{-12} , C^2 \]
Calculate \( r^2 \):
\[ (0.002)^2 = 4 \times 10^{-6} , m^2 \]
Now plug these into the equation:
\[ F_e = (8.99 \times 10^9) \frac{18 \times 10^{-12}}{4 \times 10^{-6}} \]
\[ F_e = (8.99 \times 10^9) \frac{18}{4} \times 10^{-6} \]
\[ F_e = (8.99 \times 10^9) \times 4.5 \times 10^{-6} \]
Now calculating:
\[ F_e \approx 8.99 \times 4.5 \times 10^3 \]
\[ F_e \approx 40.455 \times 10^3 , N \] \[ F_e \approx 4.0455 \times 10^4 , N \approx 4 \times 10^4 , N \]
Now for the direction: The negative charge \( q_1 \) attracts the positive charge \( q_2 \). Since \( q_2 \) is to the south of \( q_1 \), the direction of the force on \( q_2 \) will be towards \( q_1 \), which is downwards (south).
Thus, the final answer is:
Magnitude: \( 4 \times 10^{4} , N \)
Direction: South.
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