two point charges are placed on the x axis as follows:
charge q1 +4.00nc is located at x = 0.200m, and charge q2 = +5.00nc is at x = -0.300m
what are the magnitude and directionof the net force exerted by these two charges on a negative point charge q3 = -0.600nc placed at the origin ?
F=240N
89
To find the magnitude and direction of the net force exerted on charge q3, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The equation is given by:
F = (k * |q1 * q3|) / r^2
Where:
- F is the force between the charges
- k is the electrostatic constant (9.0 x 10^9 N*m^2/C^2)
- q1 and q3 are the charges
- r is the distance between the charges
Let's calculate the force between charge q1 and q3 first:
q1 = +4.00 nC (positive charge)
q3 = -0.60 nC (negative charge)
r1 = 0.2 m (distance between q1 and q3)
Using Coulomb's Law:
F1 = (k * |q1 * q3|) / r1^2
F1 = (9.0 x 10^9 N*m^2/C^2 * |+4.00 nC * -0.60 nC|) / (0.2 m)^2
F1 = (9.0 x 10^9 * 2.40 nC^2) / 0.04 m^2
F1 = 540.0 N
The force between charge q1 and q3 is 540.0 N.
Now, let's calculate the force between charge q2 and q3:
q2 = +5.00 nC (positive charge)
r2 = 0.3 m (distance between q2 and q3)
Using Coulomb's Law:
F2 = (k * |q2 * q3|) / r2^2
F2 = (9.0 x 10^9 N*m^2/C^2 * |+5.00 nC * -0.60 nC|) / (0.3 m)^2
F2 = (9.0 x 10^9 * 3.00 nC^2) / 0.09 m^2
F2 = 2700 N
The force between charge q2 and q3 is 2700 N.
To find the net force, we need to add the forces vectorially. Since F1 is positive and F2 is negative (due to opposite charges), we can subtract their magnitudes to find the net force:
Net Force = |F1| - |F2|
Net Force = 540.0 N - 2700 N
Net Force = -2160 N
The magnitude of the net force exerted by the two charges on charge q3 is 2160 N. The negative sign indicates that the force is directed in the opposite direction to the positive x-axis.
Therefore, the net force is 2160 N, and it is directed in the negative x-axis direction.
To find the magnitude and direction of the net force exerted by the two charges on the negative point charge, we can use Coulomb's Law.
Coulomb's Law 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. Mathematically, it can be expressed as:
F = k * |q1*q2| / r^2
Where:
- F is the magnitude of the force
- k is the electrostatic constant (k = 8.99 * 10^9 Nm^2/C^2)
- q1 and q2 are the charges
- r is the distance between the charges
In this case, we have:
- q1 = +4.00 nC
- q2 = +5.00 nC
- q3 = -0.600 nC
- Distance between q1 and q3 (r13) = 0.200 m
- Distance between q2 and q3 (r23) = 0.300 m
Now, let's calculate the force exerted on q3 by each charge separately and find the net force.
Force exerted by q1 on q3:
F13 = k * |q1*q3| / r13^2
F13 = (8.99 * 10^9 Nm^2/C^2) * |(4.00 * 10^-9 C) * (-0.600 * 10^-9 C)| / (0.200 m)^2
Calculating F13 will give you the magnitude of the force exerted by q1 on q3. The negative sign in front of q3 indicates that the force will be attractive.
Force exerted by q2 on q3:
F23 = k * |q2*q3| / r23^2
F23 = (8.99 * 10^9 Nm^2/C^2) * |(5.00 * 10^-9 C) * (-0.600 * 10^-9 C)| / (0.300 m)^2
Calculating F23 will give you the magnitude of the force exerted by q2 on q3. The negative sign in front of q3 indicates that the force will also be attractive.
To find the net force, we can add the vectors representing F13 and F23, taking into account their magnitudes and directions. Since the forces are attractive, their magnitudes will be added, and the direction will be towards the negative charge q3.
Net force magnitude: Fnet = |F13| + |F23|
Net force direction: Towards the negative charge q3.
By calculating the values of F13 and F23, you can find the net force exerted by these two charges on the negative point charge q3.