A sphere of radiusRhas a uniform of volume charge density. The electric potential at a pointr,r

Question

A sphere of radiusRhas a uniform of volume charge density. The electric potential at a pointr,r<R isa)due to the charge in the spherical shell of inner and outer radiirandRonlyb)Independent ofrc)due to the charge inside a sphere of radiusronlyd)due to entire charge of the sphereCorrect answer is option 'C'. Can you explain this answer?

Answer

To find the electric potential at a point (r, θ, φ) inside a sphere of radius R with a uniform volume charge density, we can use the formula for the electric potential created by a continuous charge distribution:

V = k ∫ (ρ / r) dV

where V is the electric potential, k is the Coulomb's constant (k = 1 / (4πε₀)), ρ is the charge density, r is the distance from the charge element to the point where the potential is being calculated, and dV is the differential volume element.

In this case, since the charge density is uniform, ρ is a constant. Therefore, we can write:

V = k ∫ (ρ / r) dV
= k ρ ∫ dV / r

The integral is taken over the volume of the sphere. Since the sphere has spherical symmetry, we can use spherical coordinates to simplify the integral. The differential volume element in spherical coordinates is given by:

dV = r² sinθ dr dθ dφ

Substituting this into the integral:

V = k ρ ∫ r² sinθ dr dθ dφ / r
= k ρ ∫ r sinθ dr dθ dφ

The limits of integration for r, θ, and φ will depend on the region of the sphere being considered. However, since the question does not specify a specific region, we will assume that we are considering the entire sphere.

The limits of integration for r will be from 0 to R, θ will be from 0 to π, and φ will be from 0 to 2π.

V = k ρ ∫₀²π ∫₀ᴨ ∫₀ʳ r sinθ dr dθ dφ
= k ρ ∫₀²π ∫₀ᴨ [-cosθ]₀ʳ dθ dφ
= k ρ ∫₀²π [-cosθ]₀ᴨ dφ
= k ρ ∫₀²π (1 - (-1)) dφ
= k ρ ∫₀²π 2 dφ
= 2k ρ φ ∣₀²π
= 2k ρ (2π - 0)
= 4πk ρ

Therefore, the electric potential at a point (r, θ, φ) inside the sphere is 4πkρ.

Answer

L did ans d

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