The magnitude of the charges is |q| = 7.8 µC, and the distance between them is d = 2.7 mm.
Charge 2 is attracted to charge 1, because they are opposite charges. Charge 2 is repelled from charge 3, because they are like charges. Therefore, both 21 and 23 point to the left.
F=?
I have no idea what the geometric arrangement is, however, add the forces as vectors.
To calculate the force between charges 1 and 2 (F12) and between charges 2 and 3 (F23), you can use Coulomb's Law, which states that the magnitude of the electrostatic force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
Coulomb's Law equation is given as:
F = k * (|q1| * |q2|) / r^2
Where:
- F is the magnitude of the electrostatic force
- k is the Coulomb's constant (k = 8.99 x 10^9 Nm^2/C^2)
- |q1| and |q2| are the magnitudes of the two charges
- r is the distance between the charges
Let's calculate the forces:
1. Force between charges 1 and 2 (F12):
- |q1| = 7.8 µC (This is the magnitude of charge 1)
- |q2| = 7.8 µC (This is the magnitude of charge 2)
- r = 2.7 mm (Convert to meters by dividing by 1000: r = 2.7 / 1000 = 0.0027 m)
Plugging these values into the equation, we have:
F12 = (8.99 x 10^9 Nm^2/C^2) * (7.8 x 10^-6 C) * (7.8 x 10^-6 C) / (0.0027 m)^2
2. Force between charges 2 and 3 (F23):
- |q1| = 7.8 µC (This is the magnitude of charge 2)
- |q2| = 7.8 µC (This is the magnitude of charge 3)
- r = 2.7 mm (Convert to meters by dividing by 1000: r = 2.7 / 1000 = 0.0027 m)
Plugging these values into the equation, we have:
F23 = (8.99 x 10^9 Nm^2/C^2) * (7.8 x 10^-6 C) * (7.8 x 10^-6 C) / (0.0027 m)^2
By calculating these equations, you will get the magnitudes of the forces F12 and F23 between the charges 1 and 2, and 2 and 3 respectively.