The drawing shows a parallel plate capacitor that is moving with a speed of 30 m/s through a 3.1 T magnetic field. The velocity vector v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 170 N/C, and each plate has an area of 7.5 10-4 m2. What is the magnetic force (magnitude and direction) exerted on the positive plate of the capacitor?

E = Q/εoA, => Q = εo AE

Q = 220(8.85x10-12)(8x10-4) = 1.5576 x10^-12 C

F = q v B sin θ = 1.56(26)3.1E-12 = 125.7E -12 N

To find the magnetic force exerted on the positive plate of the capacitor, we can use the equation for the magnetic force on a moving charged particle:

F = q * v * B

Where:
F is the magnetic force
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field

In this case, the charge of the particle is the charge on the positive plate of the capacitor. The velocity and the magnetic field are given in the problem.

Given:
v = 30 m/s
B = 3.1 T
E = 170 N/C
A = 7.5 * 10^-4 m^2 (area of each plate)

The electric field, E, is related to the charge density, σ, on the plates of the capacitor:

E = σ / ε₀

Where:
ε₀ is the permittivity of free space.

We can rearrange this equation to solve for σ:

σ = E * ε₀

Given that ε₀ = 8.85 * 10^-12 F/m, we can calculate σ:

σ = (170 N/C) * (8.85 * 10^-12 F/m) = 1.5045 * 10^-9 C/m^2

The charge, q, on the positive plate can be found using the formula:

q = σ * A

q = (1.5045 * 10^-9 C/m^2) * (7.5 * 10^-4 m^2) = 1.128375 * 10^-12 C

Now we can calculate the magnetic force using the formula:

F = q * v * B

F = (1.128375 * 10^-12 C) * (30 m/s) * (3.1 T)

F = 1.0526075 * 10^-10 N

Therefore, the magnitude of the magnetic force exerted on the positive plate of the capacitor is approximately 1.05 * 10^-10 N.

Since the velocity vector is perpendicular to the magnetic field, the direction of the magnetic force can be determined by applying the right-hand rule. In this case, it would be perpendicular to both the velocity vector and the magnetic field, coming out of the page or screen if the diagram is represented on one.

To find the magnetic force exerted on the positive plate of the capacitor, we can use the equation for the magnetic force on a moving charge:

F = q * v * B

where:
F is the magnetic force,
q is the charge of the positive plate of the capacitor,
v is the velocity of the capacitor,
B is the magnetic field strength.

In this case, we are given:
v = 30 m/s (velocity of the capacitor)
B = 3.1 T (magnetic field strength)

To calculate the charge of the positive plate, we can use the equation:

q = E * A

where:
q is the charge,
E is the electric field strength within the capacitor,
A is the area of each plate.

In this case, we are given:
E = 170 N/C (electric field within the capacitor)
A = 7.5 * 10^-4 m^2 (area of each plate)

Now, let's calculate the charge of the positive plate:

q = E * A
= 170 N/C * 7.5 * 10^-4 m^2
= 0.1275 C

Substituting the values into the equation for the magnetic force, we have:

F = q * v * B
= 0.1275 C * 30 m/s * 3.1 T

Calculating the value:

F ≈ 11.9125 N

So, the magnetic force exerted on the positive plate of the capacitor is approximately 11.9125 N.

Since the charge of the positive plate is positive, and the velocity and magnetic field are perpendicular to each other, the magnetic force will act in a direction perpendicular to both the magnetic field and the velocity. According to the right-hand rule, the force will be perpendicular to the plane containing the velocity vector and the magnetic field vector.