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To solve this problem, we need to find the individual forces exerted on charge [tex]\( q_3 \)[/tex] by charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex], and then compute the net force on [tex]\( q_3 \)[/tex]. Here are the given values:
- Charge [tex]\( q_1 = 2.0 \times 10^{-6} \)[/tex] C
- Charge [tex]\( q_2 = -3.0 \times 10^{-6} \)[/tex] C
- Charge [tex]\( q_3 = 1.21 \times 10^{-6} \)[/tex] C
- Distance between [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex], [tex]\( r_{13} = 0.05 \)[/tex] m
- Distance between [tex]\( q_2 \)[/tex] and [tex]\( q_3 \)[/tex], [tex]\( r_{23} = 0.03 \)[/tex] m
Coulomb's constant: [tex]\( k = 8.99 \times 10^9 \)[/tex] N m[tex]\(^2\)[/tex] C[tex]\(^{-2}\)[/tex]
### Step-by-Step Solution
1. Calculate the force [tex]\( \vec{F}_1 \)[/tex] exerted by [tex]\( q_1 \)[/tex] on [tex]\( q_3 \)[/tex]:
The formula for the electrostatic force [tex]\( F \)[/tex] is given by Coulomb’s law:
[tex]\[ F = k \frac{|q_1 q_3|}{r_{13}^2} \][/tex]
Plugging in the values:
[tex]\[ F_1 = 8.99 \times 10^9 \frac{(2.0 \times 10^{-6} \times 1.21 \times 10^{-6})}{(0.05)^2} \][/tex]
By computing this value, we get:
[tex]\[ F_1 = 8.702319999999999 \text{ N} \][/tex]
Since both charges [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex] are positive, the force [tex]\( F_1 \)[/tex] is repulsive and directed to the right, so [tex]\( \vec{F}_1 \)[/tex] is positive.
2. Calculate the force [tex]\( \vec{F}_2 \)[/tex] exerted by [tex]\( q_2 \)[/tex] on [tex]\( q_3 \)[/tex]:
Again, using Coulomb’s law:
[tex]\[ F_2 = k \frac{|q_2 q_3|}{r_{23}^2} \][/tex]
Plugging in the values:
[tex]\[ F_2 = 8.99 \times 10^9 \frac{(3.0 \times 10^{-6} \times 1.21 \times 10^{-6})}{(0.03)^2} \][/tex]
By computing this value, we get:
[tex]\[ F_2 = 36.25966666666666 \text{ N} \][/tex]
Since [tex]\( q_2 \)[/tex] is negative and [tex]\( q_3 \)[/tex] is positive, the force [tex]\( F_2 \)[/tex] is attractive and hence also directed to the right, so [tex]\( \vec{F}_2 \)[/tex] is positive.
3. Calculate the net force [tex]\( \vec{F} \)[/tex] on [tex]\( q_3 \)[/tex]:
The net force is the sum of the individual forces [tex]\( \vec{F}_1 \)[/tex] and [tex]\( \vec{F}_2 \)[/tex]:
[tex]\[ F_{\text{net}} = F_1 + F_2 \][/tex]
Substituting the values:
[tex]\[ F_{\text{net}} = 8.702319999999999 \text{ N} + 36.25966666666666 \text{ N} \][/tex]
By adding these values, we get:
[tex]\[ F_{\text{net}} = 44.96198666666666 \text{ N} \][/tex]
Therefore, the net force [tex]\( \vec{F} \)[/tex] on [tex]\( q_3 \)[/tex] is [tex]\( 44.96198666666666 \)[/tex] N directed to the right.
In summary:
[tex]\[ \vec{F}_1 = 8.702319999999999 \text{ N} \quad \text{(to the right)} \][/tex]
[tex]\[ \vec{F}_2 = 36.25966666666666 \text{ N} \quad \text{(to the right)} \][/tex]
[tex]\[ \vec{F}_{\text{net}} = 44.96198666666666 \text{ N} \quad \text{(to the right)} \][/tex]
- Charge [tex]\( q_1 = 2.0 \times 10^{-6} \)[/tex] C
- Charge [tex]\( q_2 = -3.0 \times 10^{-6} \)[/tex] C
- Charge [tex]\( q_3 = 1.21 \times 10^{-6} \)[/tex] C
- Distance between [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex], [tex]\( r_{13} = 0.05 \)[/tex] m
- Distance between [tex]\( q_2 \)[/tex] and [tex]\( q_3 \)[/tex], [tex]\( r_{23} = 0.03 \)[/tex] m
Coulomb's constant: [tex]\( k = 8.99 \times 10^9 \)[/tex] N m[tex]\(^2\)[/tex] C[tex]\(^{-2}\)[/tex]
### Step-by-Step Solution
1. Calculate the force [tex]\( \vec{F}_1 \)[/tex] exerted by [tex]\( q_1 \)[/tex] on [tex]\( q_3 \)[/tex]:
The formula for the electrostatic force [tex]\( F \)[/tex] is given by Coulomb’s law:
[tex]\[ F = k \frac{|q_1 q_3|}{r_{13}^2} \][/tex]
Plugging in the values:
[tex]\[ F_1 = 8.99 \times 10^9 \frac{(2.0 \times 10^{-6} \times 1.21 \times 10^{-6})}{(0.05)^2} \][/tex]
By computing this value, we get:
[tex]\[ F_1 = 8.702319999999999 \text{ N} \][/tex]
Since both charges [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex] are positive, the force [tex]\( F_1 \)[/tex] is repulsive and directed to the right, so [tex]\( \vec{F}_1 \)[/tex] is positive.
2. Calculate the force [tex]\( \vec{F}_2 \)[/tex] exerted by [tex]\( q_2 \)[/tex] on [tex]\( q_3 \)[/tex]:
Again, using Coulomb’s law:
[tex]\[ F_2 = k \frac{|q_2 q_3|}{r_{23}^2} \][/tex]
Plugging in the values:
[tex]\[ F_2 = 8.99 \times 10^9 \frac{(3.0 \times 10^{-6} \times 1.21 \times 10^{-6})}{(0.03)^2} \][/tex]
By computing this value, we get:
[tex]\[ F_2 = 36.25966666666666 \text{ N} \][/tex]
Since [tex]\( q_2 \)[/tex] is negative and [tex]\( q_3 \)[/tex] is positive, the force [tex]\( F_2 \)[/tex] is attractive and hence also directed to the right, so [tex]\( \vec{F}_2 \)[/tex] is positive.
3. Calculate the net force [tex]\( \vec{F} \)[/tex] on [tex]\( q_3 \)[/tex]:
The net force is the sum of the individual forces [tex]\( \vec{F}_1 \)[/tex] and [tex]\( \vec{F}_2 \)[/tex]:
[tex]\[ F_{\text{net}} = F_1 + F_2 \][/tex]
Substituting the values:
[tex]\[ F_{\text{net}} = 8.702319999999999 \text{ N} + 36.25966666666666 \text{ N} \][/tex]
By adding these values, we get:
[tex]\[ F_{\text{net}} = 44.96198666666666 \text{ N} \][/tex]
Therefore, the net force [tex]\( \vec{F} \)[/tex] on [tex]\( q_3 \)[/tex] is [tex]\( 44.96198666666666 \)[/tex] N directed to the right.
In summary:
[tex]\[ \vec{F}_1 = 8.702319999999999 \text{ N} \quad \text{(to the right)} \][/tex]
[tex]\[ \vec{F}_2 = 36.25966666666666 \text{ N} \quad \text{(to the right)} \][/tex]
[tex]\[ \vec{F}_{\text{net}} = 44.96198666666666 \text{ N} \quad \text{(to the right)} \][/tex]
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