What is the mass of a sphere with uniform charge?

[ghost34]

The sphere has radius R, and uniform volume charge density P. This sphere remains stationary (levitates) when placed above an infinite sheet of paper with a uniform surface charge density u. What is this sphere's mass?

 

[voko]

The sphere can be replaced with a point charge (the charge and mass of the sphere).

 

[Blibbler]

The weight of the sphere is equal to the electrostatic force between the sphere and the sheet when the sphere is levitating. Because the sheet is infinite the sphere is infinitessimal and is therefore a point. Write the equation for the electrostatic force and for the gravitational force and rearrange to give mass on one side of the equation and everything else on the other.

 

[vanhees71]

Be careful not to forget to take into account influence (hint: can be calculated by the method of mirror charges).

 

[sardar]

The mass of the sphere can be calculated using the formula for the mass of a charged object, which is given by M = (4/3)πR^3P, where R is the radius and P is the volume charge density. In this case, we know that the sphere has a uniform charge distribution, so the volume charge density can be expressed as P = Q/V, where Q is the total charge and V is the volume of the sphere.

To find the total charge, we can use the formula Q = uA, where u is the surface charge density and A is the area of the sphere. The area of a sphere is given by A = 4πR^2. Substituting this into the formula for Q, we get Q = u(4πR^2).

Now, we can substitute this value for Q into the formula for the mass of the sphere, giving us M = (4/3)πR^3(u(4πR^2))/V.

Since the sphere remains stationary when placed above the infinite sheet of paper, we can assume that the gravitational force acting on the sphere is equal to the electric force. This means that the weight of the sphere is balanced by the electric force, which is given by F = QEu, where E is the electric field strength. Since the sphere is stationary, we can equate the two forces, giving us QEu = Mg, where g is the acceleration due to gravity.

Solving for M, we get M = QEu/g. Substituting our value for Q, we get M = u(4πR^2)Eu/g. Finally, we can substitute the formula for the electric field strength above the infinite sheet of paper, E = u/2ε0, where ε0 is the permittivity of free space, giving us M = (u^2)(4πR^2)/2ε0g.

Therefore, the mass of the sphere is given by M = (u^2)(4πR^2)/2ε0g. This formula shows that the mass of the sphere is directly proportional to the square of the surface charge density and the radius of the sphere, and inversely proportional to the acceleration due to gravity and the permittivity of free space.