An insulating hollow sphere has inner radius a and outer radius b. Within the insulating material the volume charge density is given by rho(r)=αr, where α is a positive constant. A). What is the magnitude of the electric field at a distance r from the center of the shell, where a Express your answer in terms of the variables α, a, r, and electric constant ϵ0.
B) .A point charge
q is placed at the center of the hollow space, at r=0. What value must q have (sign and magnitude) in order for the electric field to be constant in the region a Express your answer in terms of the variables α, a, and appropriate constants.
C). What then is the value of the constant field in this region?
Express your answer in terms of the variable αand electric constant ϵ0.

E   = α/2∈₀ [ 1 - a²/r² ]

Ф = α/2∈₀

Explanation:

Using Gauss Law:

    ρ(r) = a/r, dA

          = 4 π r²d r

    Ф = ρ(r')dA

    Ф = ρ(r')dA

             = 4πα r'dr'

Ф     = 4 π α 1/2(r²-a²)

E(4πr²) = ∈₀

           = ∈₀(4πr²)

           = α (r² - a²) / 2 ∈₀ (r²)

           = α/2∈₀ [ r²/r² - a²/r² ]

      E   = α/2∈₀ [ 1 - a²/r² ]

Electric field of the point charge:

E = q / 4π∈₀r²

= α / 2 ∈₀ - (α / 2 ∈₀ )(a² / r²) + q / 4 π ∈₀ r²

For  to be constant:

- (αa²/ 2 ∈₀ ) + q / 4 π ∈₀ = 0 and q = 2παa²

-> α / 2 ∈₀ - αa²/ 2 ∈₀ + 2παa² / 4 π ∈₀

= α - αa² + αa² / 2 ∈₀

= α /2 ∈₀

Hence:

Ф = α/2∈₀

increasing transnational kinetic energy. because the transnational kinetic energy formula consists of only two variables, mass and velocity, increasing one of those properties is the only way to increase an object's transnational kinetic energy. increases to mass and velocity, however, do not have the same impact.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards