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Question:

A solid ball of radius R has a charge density ρ given by ρ = ρ₀(1−r/R) for 0 ≤ r ≤ R. The electric field outside the ball is:

ρ₀R³/12 ε₀r²

ρ₀R³/ε₀r²

4ρ₀R³/3ε₀r²

3 ρ₀R³/4ε₀r²

Solution:

ρ = ρ₀(1−r/R)
dq = ρdv
qin = ∫dq = ∫ρdv
ρ₀(1−r/R)4πr²dr
∫dv = 4πr²dr
= 4πρ₀∫₀ᴿ(1−r/R)r²dr
= 4πρ₀∫₀ᴿr²dr − ∫₀ᴿr³/R dr
= 4πρ₀[[r³/3]₀ᴿ − [r⁴/4R]₀ᴿ]
= 4πρ₀[R³/3 − R³/4]
= 4πρ₀[R³/12]
q = πρ₀R³/3
E.4πr² = (πρ₀R³/3ε₀) ⇒ E = ρ₀R³/12ε₀r²