The electrostatic force on charge 3 is 16.9 N to the right
Explanation:
The magnitude of the electrostatic force between two charges is given by Coulomb's law:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the two charges
r is the separation between the two charges
The force is attractive if the two charges have opposite sign and repulsive if the two charges have same sign.
Here we have to find the total force on the 3rd charge, so we have to calculate the vector sum of the forces produced by the other 2 charges.
For the force due to charge 1, we have
[tex]q_1=-7 \mu C = -7\cdot 10^{-6}C\\q_3 = -6 \mu C=-6\cdot 10^{-6}C[/tex]
Charge 1 is located at x = 10 cm while charge 3 is at x = -2 cm, so their distance is
[tex]d=10-(-2)=12 cm = 0.12 m[/tex]
So the force between them is
[tex]F_{13}=k\frac{q_1 q_3}{r^2}=\frac{(8.99\cdot 10^9)(-7\cdot 10^{-6})(-6\cdot 10^{-6})}{(0.12)^2}=26.3 N[/tex]
And the direction of the force on charge 3 is to the left, since the force is repulsive (charges have same sign)
For the force due to charge 2, we have
[tex]q_2=+8 \mu C = +8\cdot 10^{-6}C\\q_3 = -6 \mu C=-6\cdot 10^{-6}C[/tex]
Charge 2 is located at x = 8 cm while charge 3 is at x = -2 cm, so their distance is
[tex]r=8-(-2)=10 cm = 0.10 m[/tex]
So the force between them is
[tex]F_{23}=k\frac{q_2 q_3}{r^2}=\frac{(8.99\cdot 10^9)(8\cdot 10^{-6})(-6\cdot 10^{-6})}{(0.10)^2}=-43.2 N[/tex]
And the direction of the force on charge 3 is to the right, since the force is attractive (charges have opposite sign)
Therefore, the net force on charge 3 is
[tex]F=F_{13}+F_{23}=26.3+(-43.2)=-16.9 N[/tex]
So, the net force is 16.9 N to the right.
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