Sphere above thin conducting sheet

[irrek]

A sphere of mass m = 0.45 kg is suspended 0.20 m above a thin sheet of charge with an area of 0.65 m2 and uniform surface charge density σ. The sphere has a uniform charge density of 4 nC/m3 and a radius of 3.2x10-3 m. You may ignore adverse field effects from the edges of the sheet of charge, meaning that you may expect a uniform electric field from the conducting sheet.

1. Find σ, the surface charge density of the thin conducting sheet.
a. Find the total charge of the thin conducting sheet.

Keeping the same configuration as above, the thin conducting sheet is now replaced by a dipole made of two identical point charges fixed to the x-axis a distance d = 0.55 m apart. The position and orientation of the sphere does not change.

2.. What must the sign of each charge be to support the sphere?
3. Calculate the magnitude of each point charge.
4. Find the angle between the vector forces from each charge in the dipole
5. Report the straight line distance between each point charge and the sphere of charge.

A position vector P has a magnitude of 1.25 m and makes an angle φ = π/3 rad with the +x-axis.

6. Given the configuration of the dipole and charged sphere above, what is the electric field at point P?
7. If the magnitude of P were tripled what would happen to the magnitude of the electric field?



E = σ/(2ε0)
mg = qE

All I've been able to figure out is the very first question.
m = 0.45 kg, ρ = 4 nC/m3, r = 3.2*10-3 m,
charge on the ball q = ρ*(4πr3/3)
E = σ/(2ε0)
mg = qE
mg = ρ*(4πr3/3) * σ/(2ε0)
σ = 3ε0mg/(2πρr3) = 1.42*105 C/m2

After that, I'm completely lost
!

 

[anathema]

Hello! I would be happy to help you with the remaining questions. Let's break them down one by one and see if we can come up with some solutions.

2. To support the sphere, the charges in the dipole must have opposite signs. This is because the electric field created by one charge will repel the sphere, while the electric field created by the other charge will attract the sphere. This balance of forces will keep the sphere suspended in place.

3. To calculate the magnitude of each point charge, we can use the equation q = ρ*(4πr3/3), where q is the charge, ρ is the charge density, and r is the radius of the sphere. Plugging in the values given in the problem, we get q = (4 nC/m3)*(4π*(3.2*10-3 m)3/3) = 1.03*10-5 C. This means that each point charge in the dipole will have a magnitude of 1.03*10-5 C.

4. The angle between the vector forces from each charge in the dipole can be found using the equation F = kq1q2/r2, where F is the force, k is the Coulomb constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the two charges. Since the charges are fixed on the x-axis, the distance between them is simply d = 0.55 m. Plugging in the values, we get F = (9*109 N*m2/C2)*(1.03*10-5 C)2/(0.55 m)2 = 2.97*10-3 N. To find the angle, we can use the equation θ = tan-1(Fy/Fx), where Fy and Fx are the y-component and x-component of the force, respectively. Since the charges have opposite signs, one will have a positive y-component and the other will have a negative y-component. This means that the angle between the forces will be 180 degrees, or π radians.

5. The straight line distance between each point charge and the sphere can be found using the Pythagorean theorem. The distance between the center of the sphere and one of the point charges will be the hypotenuse of a right triangle, with the distance between the two point charges (d