Let's go through the questions step by step.
Question 2.1: State Coulomb's Law in words.
Coulomb's Law states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The force is attractive if the charges are of opposite signs and repulsive if they are of the same sign.
Question 2.2: Calculate the force of Qx on Qw.
Coulomb's Law is mathematically represented as:
\[ F = k \frac{|Q_1 Q_2|}{r^2} \]
where:
- \( F \) = electrostatic force between the charges,
- \( k \) = Coulomb's constant โ \( 8.99 \times 10^9 , \text{N m}^2/\text{C}^2 \),
- \( Q_1 \) and \( Q_2 \) = magnitudes of the charges,
- \( r \) = distance between the charges.
Given:
- \( Q_x = +5 , \text{nC} = 5 \times 10^{-9} , \text{C} \)
- \( Q_w = -14 , \mu\text{C} = -14 \times 10^{-6} , \text{C} \)
- Distance (\( r \)) = 20 cm = 0.2 m
Now plug in the values:
\[ F_{xw} = k \frac{|Q_x Q_w|}{r^2} \]
\[ F_{xw} = (8.99 \times 10^9) \frac{|(5 \times 10^{-9})(-14 \times 10^{-6})|}{(0.2)^2} \]
Calculate:
\[ F_{xw} = (8.99 \times 10^9) \frac{(5 \times 10^{-9})(14 \times 10^{-6})}{0.04} \]
\[ F_{xw} = (8.99 \times 10^9) \frac{70 \times 10^{-15}}{0.04} \]
\[ F_{xw} = (8.99 \times 10^9) \times (1.75 \times 10^{-12}) \]
\[ F_{xw} = 15.7325 , \text{N} \approx 15.733 , \text{N} \]
Question 2.3: Calculate the force of Qy on Qw.
For the force of Qy on Qw, we follow the same process:
Given:
- \( Q_y = -8 , \text{nC} = -8 \times 10^{-9} , \text{C} \) Distance from Qw to Qy is 30 cm (0.3 m).
Using Coulomb's Law:
\[ F_{yw} = k \frac{|Q_y Q_w|}{r^2} \]
\[ F_{yw} = (8.99 \times 10^9) \frac{|(-8 \times 10^{-9})(-14 \times 10^{-6})|}{(0.3)^2} \]
Calculate:
\[ F_{yw} = (8.99 \times 10^9) \frac{(8 \times 10^{-9})(14 \times 10^{-6})}{0.09} \]
\[ F_{yw} = (8.99 \times 10^9) \frac{112 \times 10^{-15}}{0.09} \]
\[ F_{yw} = (8.99 \times 10^9) \times (1.244444 \times 10^{-12}) \]
\[ F_{yw} \approx 11.202 , \text{N} \approx 11.202 , \text{N} \]
Question 2.4: Calculate the magnitude and direction of the net electrostatic force on Qw.
The forces \( F_{xw} \) and \( F_{yw} \) are perpendicular (east and south directions, respectively).
To find the resultant force:
Using Pythagorean theorem:
\[ F_{\text{net}} = \sqrt{(F_{xw})^2 + (F_{yw})^2} \]
\[ F_{\text{net}} = \sqrt{(15.733)^2 + (11.202)^2} \]
Calculate:
\[ F_{\text{net}} = \sqrt{247.20 + 125.45} \]
\[ F_{\text{net}} = \sqrt{372.65} \]
\[ F_{\text{net}} \approx 19.27 , \text{N} \]
Now, calculate the angle \( \theta \) with respect to the positive x-axis (east direction):
\[ \theta = \tan^{-1}\left(\frac{F_{yw}}{F_{xw}}\right) \]
\[ \theta = \tan^{-1}\left(\frac{11.202}{15.733}\right) \]
Calculating:
\[ \theta \approx \tan^{-1}(0.712) \approx 35.00^\circ \text{ (south of east)} \]
Question 2.5.1: Draw the electric field between the two charges.
The electric field lines would emanate from the positive charge \( Q_x \) and converge towards the negative charge \( Q_w \). Since \( Q_w \) is negative and \( Q_x \) is positive, the electric field points from \( Q_x \) towards \( Q_w \).
Question 2.5.2: Calculate how much charge was transferred when after the charges were in contact.
When \( Q_x \) and \( Q_w \) touch, they will share charge.
Total charge before contact:
\[ Q_{total} = Q_x + Q_w = (5 , \text{nC}) + (-14 , \text{ยตC}) \]
Convert \( -14 , \mu\text{C} \) to nC:
\[ -14 , \mu\text{C} = -14000 , \text{nC} \]
Now:
\[ Q_{total} = 5 + (-14000) = -13995 , \text{nC} \]
Since both charges are equal after touching, the total charge is divided equally:
\[ Q_{final} = \frac{Q_{total}}{2} = \frac{-13995}{2} = -6997.5 , \text{nC} \]
Charge transferred from \( Q_x \) to \( Q_w \):
Charge initially with \( Q_x \) is +5 nC.
So, charge transferred:
\[ \text{Charge transferred} = Q_{\text{final}} - Q_x = -6997.5 - 5 = -7002.5 , \text{nC} \]
So approximately:
\[ \text{Charge transferred} \approx -7002.5 , \text{nC} \]
Summary of Answers:
- Q2.1: Coulomb's Law describes the relationship between the force between two charges based on their magnitudes and the distance separating them.
- Q2.2: \( F_{xw} \approx 15.733 , \text{N} \)
- Q2.3: \( F_{yw} \approx 11.202 , \text{N} \)
- Q2.4: \( F_{\text{net}} \approx 19.27 , \text{N} \) at \( 35.00^\circ \) south of east.
- Q2.5.1: Electric field diagram (drawing needed).
- Q2.5.2: Charge transferred โ \(-7002.5 , \text{nC}\)
Feel free to ask if you need further assistance!
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