a charge of 7.2 nC and a charge of 6.7 nC are separated b 32 cm. Find the equilibrium position for a -3.0 nC charge.
9.06 or 0.16
First, see if you can get an equilibrium postition between the two charges.
Figure the force from each of the two charges, and set them equal (they will be in opposing sides, so this is valid).
K*7.2*3.0/x^2 = k*6.7*3.2/(32-x)^2
find x
I discarded the negative signs as they are on both sides, put them in if you wish.
To find the equilibrium position for a -3.0 nC charge, we can use Coulomb's Law which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
1. Let's denote the first charge as q1 = 7.2 nC, the second charge as q2 = 6.7 nC, and the distance between them as r = 32 cm = 0.32 m.
2. Now, we need to find the force between these two charges using Coulomb's Law. The formula for Coulomb's Law is given as:
F = (k * |q1 * q2|) / r^2
where F is the force, k is Coulomb's constant (9 × 10^9 N·m^2/C^2), q1 and q2 are the charges, and r is the distance between them.
Plugging in the values:
F = (9 × 10^9 N·m^2/C^2 * |7.2 nC * 6.7 nC|) / (0.32 m)^2
Note: The absolute value is taken to ensure that the force is positive since we are only concerned with its magnitude.
3. Simplifying the expression:
F = (9 × 10^9 N·m^2/C^2 * 7.2 × 10^-9 C * 6.7 × 10^-9 C) / 0.32^2 m^2
F = (3.7032 N·m^2/C^2) / 0.1024 m^2
F ≈ 36.1328 N
4. The force calculated above represents the attractive force between the charges. Since the force on the -3.0 nC charge would be repulsive, the equilibrium position occurs when the repulsive force is equal in magnitude to the attractive force between the two charges.
5. So, we can set up the equation:
k * |q * q2| / r^2 = k * |q1 * q2| / r^2
where q is the charge we want to find the equilibrium position for.
Plugging in the known values:
(9 × 10^9 N·m^2/C^2 * |-3.0 nC * 6.7 nC|) / r^2 = (9 × 10^9 N·m^2/C^2 * 7.2 nC * 6.7 nC) / (0.32 m)^2
6. Simplifying the expression:
(3.603 N·m^2/C^2) / r^2 = (3.7032 N·m^2/C^2) / 0.1024 m^2
(3.603 N·m^2/C^2) / r^2 = 36.1328 N / 0.1024
7. Solving for r^2:
r^2 = (3.603 N·m^2/C^2 * 0.1024 m^2) / 36.1328 N
r^2 ≈ 0.01016 m^2
8. Taking the square root of both sides, we find:
r ≈ 0.1008 m
9. Therefore, the equilibrium position for a -3.0 nC charge is approximately 0.1008 meters away from the other charges.
To find the equilibrium position for a -3.0 nC charge, we can use Coulomb's Law, which states that the force between two charges is inversely proportional to the square of the distance between them.
First, let's determine the force between the two given charges (7.2 nC and 6.7 nC) using Coulomb's Law.
Coulomb's Law formula: F = k * (q1 * q2) / r^2
Where:
F is the force between the charges,
k is Coulomb's constant (9 * 10^9 N m^2/C^2),
q1 and q2 are the magnitudes of the two charges, and
r is the distance between the charges.
Let's calculate the force between the two charges:
F = (9 * 10^9 N m^2/C^2) * ((7.2 * 10^-9 C) * (6.7 * 10^-9 C)) / (0.32 m)^2
F ≈ 1.5 N
Now we know that the force between the charges is 1.5 N.
The force between charges is always attractive for opposite charges, and it is repulsive for charges of the same type. In this case, since the -3.0 nC charge is negative and the other charges are positive, the force between them is attractive.
To find the equilibrium position, we need to determine the position where the force acting on the -3.0 nC charge is equal in magnitude and opposite in direction to the force between the two given charges.
Since we know the force between the two given charges is 1.5 N, we need to find the position where the force on the -3.0 nC charge is also 1.5 N.
Let's use Coulomb's Law to find the distance at which the force between the -3.0 nC charge and the positive charges is 1.5 N.
1.5 N = (9 * 10^9 N m^2/C^2) * ((-3.0 * 10^-9 C) * (6.7 * 10^-9 C)) / r^2
Simplifying the equation, we have:
r^2 = ((-3.0 * 10^-9 C) * (6.7 * 10^-9 C)) / (1.5 N / (9 * 10^9 N m^2/C^2))
r ≈ sqrt( (18 * 10^-18 C^2) / (1.5 N / (9 * 10^9 N m^2/C^2)) )
r ≈ 0.106 m
Therefore, the equilibrium position for the -3.0 nC charge is approximately 0.106 meters away from the other charges.