Sure, let’s go through the step-by-step solution for finding [tex]\( F_1 \)[/tex], [tex]\( F_2 \)[/tex], and the net force on charge [tex]\( q_3 \)[/tex].
Given:
- The force [tex]\( \overrightarrow{F_1} \)[/tex] on [tex]\( q_3 \)[/tex] is [tex]\( 33.5 \, \text{N} \)[/tex].
- The charge [tex]\( q_3 \)[/tex] is [tex]\( -75.8 \times 10^{-6} \, \text{C} \)[/tex].
- We are provided with a value for [tex]\( q_2 \)[/tex] as [tex]\( 1 \times 10^{-6} \, \text{C} \)[/tex].
- The distance between [tex]\( q_2 \)[/tex] and [tex]\( q_3 \)[/tex] is [tex]\( 1 \, \text{m} \)[/tex].
First, let’s acknowledge the given force:
- [tex]\( \overrightarrow{F_1} \)[/tex] is [tex]\( 33.5 \, \text{N} \)[/tex].
Next, we calculate the force [tex]\( \overrightarrow{F_2} \)[/tex] exerted by charge [tex]\( q_2 \)[/tex] on [tex]\( q_3 \)[/tex].
Using Coulomb’s Law:
[tex]\[ F_2 = k \frac{|q_2 \cdot q_3|}{r^2} \][/tex]
Where:
- [tex]\( k \)[/tex] is Coulomb's constant ([tex]\( 8.988 \times 10^9 \, \text{N} \cdot \text{m}^2 \cdot \text{C}^{-2} \)[/tex]).
- [tex]\( q_2 = 1 \times 10^{-6} \, \text{C} \)[/tex].
- [tex]\( q_3 = -75.8 \times 10^{-6} \, \text{C} \)[/tex].
- [tex]\( r = 1 \, \text{m} \)[/tex].
So:
[tex]\[ F_2 = 8.988 \times 10^9 \cdot \frac{|1 \times 10^{-6} \cdot (-75.8 \times 10^{-6})|}{1^2} \][/tex]
[tex]\[ F_2 = 8.988 \times 10^9 \cdot \frac{75.8 \times 10^{-12}}{1} \][/tex]
[tex]\[ F_2 = 8.988 \times 75.8 \times 10^{-3} \][/tex]
[tex]\[ F_2 = 0.6812904 \, \text{N} \][/tex]
Since [tex]\( q_2 \)[/tex] and [tex]\( q_3 \)[/tex] have opposite charges, the force between them is attractive. Given that forces directed to the left are negative, we consider [tex]\( \overrightarrow{F_2} \)[/tex] to be [tex]\( -0.6812904 \, \text{N} \)[/tex].
Now, let's calculate the net force [tex]\( \overrightarrow{F} \)[/tex] on [tex]\( q_3 \)[/tex]:
[tex]\[ \overrightarrow{F} = \overrightarrow{F_1} + \overrightarrow{F_2} \][/tex]
[tex]\[ \overrightarrow{F} = 33.5 \, \text{N} + (-0.6812904 \, \text{N}) \][/tex]
[tex]\[ \overrightarrow{F} = 33.5 \, \text{N} - 0.6812904 \, \text{N} \][/tex]
[tex]\[ \overrightarrow{F} = 32.8187096 \, \text{N} \][/tex]
Thus, the forces are:
- [tex]\( \overrightarrow{F_1} = 33.5 \, \text{N} \)[/tex]
- [tex]\( \overrightarrow{F_2} = -0.6812904 \, \text{N} \)[/tex]
And the net force on [tex]\( q_3 \)[/tex] is:
[tex]\[ \overrightarrow{F} = 32.8187096 \, \text{N} \][/tex]
