Yes, a very sloppily worded question!Steve4Physics said:But there seems to be a mistake in the question. The sphere can't be both 'uniformly charged' and have a charge density which is a function of r.
The expression for the electric field outside a sphere is given by E = kQ/r2, where k is the Coulomb's constant, Q is the charge of the sphere, and r is the distance from the center of the sphere to the point where the electric field is being measured.
The electric field outside a sphere is inversely proportional to the square of the distance from the center of the sphere. This means that as the distance increases, the electric field decreases.
Yes, the electric field outside a sphere is directly proportional to the charge of the sphere. This means that as the charge increases, the electric field also increases.
Yes, the expression for electric field outside a sphere is valid for both positive and negative charges. However, the direction of the electric field will be different for positive and negative charges.
The electric field outside a sphere is always greater than the electric field inside the sphere. This is because the charge of the sphere is concentrated on the surface, resulting in a stronger electric field at points outside the sphere.