6. Inside a uniformly charged solid sphere of radius R, at a distance r from the centre, the
electric field is proportional to:
a. 1/r2
b. 1/r
c. r
d. r2
The correct answer is c. r. Inside a uniformly charged solid sphere, the electric field at distance r from the center is directly proportional to r.
Gauss’s Law Derivation
Gauss’s law states that the electric flux through a closed surface equals the enclosed charge divided by ε₀: ∮E⋅dA = Q_enc/ε₀. For a point inside at r < R, choose a Gaussian sphere of radius r. Symmetry makes E radial and constant on this surface, so flux = E × 4πr². The enclosed charge is the charge within volume (4/3)πr³, or Q_enc = ρ × (4/3)πr³ where ρ = Q / ((4/3)πR³) is uniform density. Thus, E = (ρ r)/(3ε₀) = (Q r)/(4π ε₀ R³), proving E ∝ r.
Option Analysis
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a. 1/r²: Applies outside (r > R), where the sphere acts like a point charge, E = Q/(4π ε₀ r²).
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b. 1/r: Incorrect; no physical scenario matches for spherical symmetry here.
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c. r: Correct inside, as derived; field is zero at center (r=0) and grows linearly to surface.
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d. r²: Wrong; would imply quadratic growth, unsupported by Gauss’s law.
The electric field inside uniformly charged solid sphere of radius R at distance r from the centre follows a linear pattern crucial for electrostatics exams. This concept, derived via Gauss’s law, shows E ∝ r for r < R, distinguishing it from external fields.
Field Variation Graphically
Inside, E increases linearly from zero at the center to maximum at r = R. Outside, it drops as 1/r², ensuring continuity at the surface.
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At r = 0: E = 0 (symmetric cancellation).
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At r = R/2: E = (Q / (8π ε₀ R²)).
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Key insight: Only enclosed charge contributes inside.
Exam Relevance
Common in CSIR NET, JEE: options test proportionality (r vs. 1/r²). Practice confirms c as answer.
