High School

Charge of [tex]97.9 \, \text{nC}[/tex] is uniformly distributed in a sphere of radius [tex]R = 12.3 \, \text{cm}[/tex]. What is the total charge contained within the radius between [tex]0[/tex] and [tex]R/2[/tex]?

Answer :

Final answer:

The total charge within the radius between 0 and R/2 of the uniformly charged sphere is about 12,24 nC. This is determined by calculating the volume ratio of the smaller sphere (with radius R/2) to the larger sphere (with radius R) and applying it to the total charge present.

Explanation:

The total charge in a sphere is uniformly distributed. If we consider a smaller sphere with a radius R/2, within the larger sphere, the charge in this smaller sphere will be proportional to its volume, compared to the larger sphere. This is due to the uniform distribution of charge.

Here is the simple calculation: We know that the volume of a sphere is 4/3πR³, so the total charge in the larger sphere (Q₁) relative to the total charge in the smaller sphere (Q₂) is the same as their volume ratios (V₁/V₂). Thus, the ratio R₁¹³/R₂³ is 1/8, where R₁ is the radius of the larger sphere and R₂ is R₁/2 is the radius of the smaller sphere. Hence Q₂, the charge contained in the radius between 0 and R/2 is Q₁/8 = 97,9 nC / 8, which is approximately 12,24 nC.

To find the total charge contained within the radius between 0 and R/2, we can use the equation for charge density on a non-uniformly charged sphere. The charge density is given by:

ρ = ∫(dq)/(dv) = Q/V = Q/(4/3πR³)

Substituting the given values, we have:

ρ = 97.9 × 10⁻⁹C/(4/3 × π × 0.123³) = 895.2 × 10³ C/m³

The charge within the radius between 0 and R/2 is then given by:

Q = ∫ρdv = ρ × (4/3π(r/2)³) = 895.2 × 10³ × (4/3 × π × (0.0615)³)

Calculating this expression will give us the total charge contained within the given radius range.

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