Question
You have two identical balls X and Y each of mass 0.20 g, carrying identical charges and suspended by two light threads of equal length from a rigid support. The balls position themselves at equilibrium such that the angle between the threads is 60°. If the distance between the balls is 0.5 m, calculate the charge on each of them. [
T
a
k
e
g
=
9.8
m
s
2
]
Question 8Select one:
1.1
×
10
-
3
C
1.8
×
10
-
7
3.1
×
10
5
C
2.3
×
10
-
6
C
T
a
k
e
g
=
9.8
m
s
2
]
Question 8Select one:
1.1
×
10
-
3
C
1.8
×
10
-
7
3.1
×
10
5
C
2.3
×
10
-
6
C
Answers
GPT-4o mini
To find the charge on each ball, we can analyze the forces acting on them when they reach equilibrium.
1. **Identify the forces:** Each ball experiences gravitational force \( F_g \) downward and electrostatic force \( F_e \) that pushes them apart due to their identical charges.
2. **Calculating gravitational force:** The mass \( m \) of each ball is given as 0.20 g, or \( 0.20 \times 10^{-3} \) kg.
\[
F_g = m \cdot g = (0.20 \times 10^{-3} \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) = 1.96 \times 10^{-3} \, \text{N}
\]
3. **Geometry of the problem:** The angle between the threads is 60°, so the angle each thread makes with the vertical is 30° (half of 60°).
With the distance between the balls being 0.5 m, we can find the horizontal distance \( x \) between the balls using trigonometry:
\[
x = 2 L \sin(30^\circ) = L \quad \text{(where \( L \) is the length of the thread)}
\]
Setting \( x = 0.5 \, m \):
\[
L = 0.5 \, m
\]
4. **Calculating horizontal distance:** The horizontal component of the tension \( T \) is what balances the electrostatic force:
\[
T_x = T \sin(30^\circ) = \frac{T}{2}
\]
The vertical component is:
\[
T_y = T \cos(30^\circ)
\]
The tension in the thread must equal the weight of the ball:
\[
T \cos(30^\circ) = F_g \implies T \cdot \frac{\sqrt{3}}{2} = 1.96 \times 10^{-3} \implies T = \frac{1.96 \times 10^{-3}}{\frac{\sqrt{3}}{2}} = \frac{1.96 \times 10^{-3} \cdot 2}{\sqrt{3}} = \frac{3.92 \times 10^{-3}}{\sqrt{3}} \approx 2.26 \times 10^{-3} \, N
\]
5. **Force balance:** Now we can equate the horizontal electrostatic force \( F_e \) to the horizontal component of the tension.
The electrostatic force between the two charges can be calculated by Coulomb's law:
\[
F_e = k \frac{q^2}{r^2}
\]
Where \( k = 8.99 \times 10^9 \, N \cdot m^2/C^2 \) and \( r = 0.5 \, m \).
Equating the two forces, we get:
\[
\frac{T}{2} = k \frac{q^2}{r^2}
\]
Substituting the values:
\[
\frac{2.26 \times 10^{-3}}{2} = (8.99 \times 10^9) \frac{q^2}{(0.5)^2}
\]
Simplifying further:
\[
1.13 \times 10^{-3} = (8.99 \times 10^9) \frac{q^2}{0.25}
\]
\[
1.13 \times 10^{-3} \cdot 0.25 = 8.99 \times 10^9 q^2
\]
\[
2.825 \times 10^{-4} = 8.99 \times 10^9 q^2
\]
Solving for \( q^2 \):
\[
q^2 = \frac{2.825 \times 10^{-4}}{8.99 \times 10^9} \approx 3.14 \times 10^{-14}
\]
Taking the square root:
\[
q \approx \sqrt{3.14 \times 10^{-14}} \approx 1.77 \times 10^{-7} \, C
\]
6. **Final answer:** Rounding appropriately gives us \( q \approx 1.8 \times 10^{-7} \, C \).
Therefore, the charge on each ball is approximately \( \boxed{1.8 \times 10^{-7} \, C} \).
1. **Identify the forces:** Each ball experiences gravitational force \( F_g \) downward and electrostatic force \( F_e \) that pushes them apart due to their identical charges.
2. **Calculating gravitational force:** The mass \( m \) of each ball is given as 0.20 g, or \( 0.20 \times 10^{-3} \) kg.
\[
F_g = m \cdot g = (0.20 \times 10^{-3} \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) = 1.96 \times 10^{-3} \, \text{N}
\]
3. **Geometry of the problem:** The angle between the threads is 60°, so the angle each thread makes with the vertical is 30° (half of 60°).
With the distance between the balls being 0.5 m, we can find the horizontal distance \( x \) between the balls using trigonometry:
\[
x = 2 L \sin(30^\circ) = L \quad \text{(where \( L \) is the length of the thread)}
\]
Setting \( x = 0.5 \, m \):
\[
L = 0.5 \, m
\]
4. **Calculating horizontal distance:** The horizontal component of the tension \( T \) is what balances the electrostatic force:
\[
T_x = T \sin(30^\circ) = \frac{T}{2}
\]
The vertical component is:
\[
T_y = T \cos(30^\circ)
\]
The tension in the thread must equal the weight of the ball:
\[
T \cos(30^\circ) = F_g \implies T \cdot \frac{\sqrt{3}}{2} = 1.96 \times 10^{-3} \implies T = \frac{1.96 \times 10^{-3}}{\frac{\sqrt{3}}{2}} = \frac{1.96 \times 10^{-3} \cdot 2}{\sqrt{3}} = \frac{3.92 \times 10^{-3}}{\sqrt{3}} \approx 2.26 \times 10^{-3} \, N
\]
5. **Force balance:** Now we can equate the horizontal electrostatic force \( F_e \) to the horizontal component of the tension.
The electrostatic force between the two charges can be calculated by Coulomb's law:
\[
F_e = k \frac{q^2}{r^2}
\]
Where \( k = 8.99 \times 10^9 \, N \cdot m^2/C^2 \) and \( r = 0.5 \, m \).
Equating the two forces, we get:
\[
\frac{T}{2} = k \frac{q^2}{r^2}
\]
Substituting the values:
\[
\frac{2.26 \times 10^{-3}}{2} = (8.99 \times 10^9) \frac{q^2}{(0.5)^2}
\]
Simplifying further:
\[
1.13 \times 10^{-3} = (8.99 \times 10^9) \frac{q^2}{0.25}
\]
\[
1.13 \times 10^{-3} \cdot 0.25 = 8.99 \times 10^9 q^2
\]
\[
2.825 \times 10^{-4} = 8.99 \times 10^9 q^2
\]
Solving for \( q^2 \):
\[
q^2 = \frac{2.825 \times 10^{-4}}{8.99 \times 10^9} \approx 3.14 \times 10^{-14}
\]
Taking the square root:
\[
q \approx \sqrt{3.14 \times 10^{-14}} \approx 1.77 \times 10^{-7} \, C
\]
6. **Final answer:** Rounding appropriately gives us \( q \approx 1.8 \times 10^{-7} \, C \).
Therefore, the charge on each ball is approximately \( \boxed{1.8 \times 10^{-7} \, C} \).