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 that is correct
What do you think? Two vectors F pointed in directions 90 degrees apart have what resultant? Think of the Pythagorean theorem. Draw a picture if necessary.
I thought it was 8 but im wrong
To find the magnitude of the net force acting on the test charge due to both of these charges, we can use the principle of superposition. The principle of superposition states that the net force on a given charge is equal to the vector sum of the individual forces exerted by each of the other charges.
In this case, we have two charges of equal magnitude (+Q and +Q) fixed at opposite corners of a square, and a test charge (-q) at a third corner. Let's consider the force acting on the test charge due to each of the other charges separately.
The magnitude of the force between two charges is given by Coulomb's law:
F = k * |q1| * |q2| / r^2
where F is the force between the charges, k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.
For the force due to the +Q charge, the distance is the length of the diagonal of the square, which is equal to a * √2 (where a is the side length of the square).
For the force due to the -Q charge, the distance is the length of one side of the square, which is equal to a.
Now, let's calculate the magnitude of the force on the test charge due to each of the other charges:
Force due to the +Q charge:
F1 = k * |-q| * |+Q| / (a * √2)^2 = k * q * Q / (2a^2)
Force due to the -Q charge:
F2 = k * |-q| * |-Q| / a^2 = k * q * Q / a^2
Since the charges are of equal magnitude, we can simplify the expressions:
F1 = F2 = k * q * Q / a^2
To find the net force, we need to sum these two forces:
Fnet = F1 + F2 = 2 * k * q * Q / a^2
Now, let's look at the answer choices and see if we can simplify the expression:
1. Fnet = 2F/√3 - This does not match our expression for Fnet.
2. Fnet = √2F - This does not match our expression for Fnet.
3. Fnet = 2F/3 - This does not match our expression for Fnet.
4. Fnet = F - This does not match our expression for Fnet.
5. Fnet = 2F - This does not match our expression for Fnet.
6. Fnet = F/√3 - This does not match our expression for Fnet.
7. Fnet = F/√2 - This does not match our expression for Fnet.
8. Fnet = 3F/2 - This does not match our expression for Fnet.
9. Fnet = 3F - This does not match our expression for Fnet.
10. Fnet = 0 - This does not match our expression for Fnet.
None of the given answer choices match our expression for the magnitude of the net force acting on the test charge. So, none of the answer choices are correct.