Two charged particles of equal magnitude
(+Q and +Q) are fixed at opposite corners of a square that lies in a plane (see figure below). A test charge −q is placed at a third corner. If F is the magnitude of the force on the test charge due to only one of the other charges,what is the magnitude of the net force acting on the test charge due to both of these
charges?
1. Fnet = 2F/√3
2. Fnet = √2F
3. Fnet = 2F/3
4. Fnet = F
5. Fnet = 2F
6. Fnet = F/√3
7. Fnet = F/√2
8. Fnet = 3F/2
9. Fnet = 3F
10. Fnet = 0
The answer is 2. Fnet=Sqrt of 2 F
To find the magnitude of the net force acting on the test charge due to both of these charges, we need to calculate the individual forces exerted by each charge and then sum them up.
The force between two charges can be calculated using Coulomb's Law, which states:
F = (k * |q1*q2|) / r^2
where F is the force, k is the Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.
In this case, the two fixed charges (+Q and +Q) are of equal magnitude, and the test charge has a magnitude of -q. Since they are fixed at opposite corners of a square, the distance between them is the length of one side of the square.
Let's denote the length of one side of the square as "d".
The force between the test charge and one of the fixed charges is then:
F = (k * |-q * Q|) / d^2
Since both fixed charges are of equal magnitude, we have:
F = (k * |-q * Q|) / d^2
To find the net force, we need to consider the forces from both fixed charges. Since the charges are at opposite corners of the square, the forces they exert on the test charge will have the same magnitude but opposite directions. This means we can simply double the force calculated above.
So, the magnitude of the net force acting on the test charge due to both charges is:
Fnet = 2 * (k * |-q * Q|) / d^2
Now that we have the expression for the net force, let's examine the answer choices and see which one matches our expression.
1. Fnet = 2F/√3 - Does not match our expression.
2. Fnet = √2F - Does not match our expression.
3. Fnet = 2F/3 - Does not match our expression.
4. Fnet = F - Does not match our expression.
5. Fnet = 2F - Matches our expression.
6. Fnet = F/√3 - Does not match our expression.
7. Fnet = F/√2 - Does not match our expression.
8. Fnet = 3F/2 - Does not match our expression.
9. Fnet = 3F - Does not match our expression.
10. Fnet = 0 - Does not match our expression.
Therefore, the correct answer is option 5. Fnet = 2F.
To find the magnitude of the net force acting on the test charge due to both of the charges, we need to calculate the individual forces acting on the test charge due to each charge and then find the vector sum.
Since the two charges, +Q and +Q, are of equal magnitude, the forces on the test charge -q due to each charge would be equal in magnitude. Let's call this force F.
Now, let's consider the forces due to each charge separately:
1. Force on the test charge due to the charge +Q: Since the charge +Q is located at the opposite corner of the square, the distance between the charge and the test charge is equal to the length of the diagonal of the square. Let's call this distance d. According to Coulomb's Law, the force between two charges is given by F = (k * q1 * q2) / r^2, where k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges. In this case, q1 = +Q, q2 = -q, and r = d. Therefore, the force on the test charge due to the charge +Q is given by F1 = (k * Q * (-q)) / d^2.
2. Force on the test charge due to the charge +Q at the other corner: Similar to the previous case, the force on the test charge due to this charge is also F, as the magnitudes and distances are the same as before.
Now, to find the net force, we need to find the vector sum of these two forces. Since both forces have the same magnitude and same direction, the resultant force would act in the same direction with the same magnitude.
Therefore, the magnitude of the net force acting on the test charge due to both of these charges is given by Fnet = 2F.
So, the correct answer is option 5: Fnet = 2F.