1E(r)=-dv/dr=-2ar,
now by gauss law, ∫E(r)ds=∫d(qin)/ε0,
now, d(qin)=4Ï€r2 Ï(r) dr,
so, -2aR*4Ï€R2=4Ï€/ε0 0∫Rr2Ï(r)dr
by newton-leibnitz formula, -8Ï€a3R2 =4Ï€/ε0 R2Ï(R)
so, Ï(R)=-6ε0a.
The field potential inside a charged ball depends only on the distance from the center as V=ar2 + b; a,b are constants. Find the space charge distribution Ï(r) inside the ball.
Src: Irodov
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7 Answers
33I'm getting two answers one in term of and other in term of b....
Ï=2bε0/R2 ...........R is radius of ball.
Ï=-6aε0
Just compared the given potential with the potential at any point at a r distance inside sphere (r<R)
that also depends on r2 that gave me a hint....
62yes dude.. Good work..
i wud say this.. the answer is there in Irodov..
I want slightly detailed proofs :)
I know this is not very tough. Atleast not at all as tough as it looks..
1
62hey i din notice abhishek had 2 expressions :D
ther is only 1 corect answer.. the one rohan gave :)
33opsie I didn't want to solve it so just compared it with cooked result for the shell.......
and compared the result at a distance (r<R).....
as the given V was also proportional to r^2 and for uniform charge distribution also V is proportional to r^2 ........ :P