推 jack8366:感謝~!我知道了 因為份部沒有平均 06/01 23:11
※ 引述《jack8366 (捲毛)》之銘言：
: ※ 引述《wyech (wicky)》之銘言：
: : 時間: Mon Jul 5 16:49:24 2010
: : 一半徑為R的非導電實心球，總電荷為Q，電合密度分佈為ρ(r)=Ar C/m3，其中A為常數。
: : 以Q來表示距中心r處的電場當r<R時
: : --
: : ◆ From: 114.41.13.44
: 我想請問一下~!這題答案真的是 Qr^2/4piεR^4 嗎?
: 因為我翻了原文書和自己算 答案都是 Qr/4piεR^3 耶....
: 有沒有同學可以幫我解一下這題...因為沒解出來害我實在是睡不著 / _> \
: 到底是我算錯了嗎....
: 我的算法是 q / (4/3)*pi*r^3 = Q / (4/3)*pi*R^3
: => q = Q(r^3/R^3)
: E*4*pi*r^2 = q / ε 把q帶入
: => E = Qr/4piεR^3
: ※ 編輯: jack8366 來自: 123.204.178.57 (05/27 10:24)
∫∫∫ρdV = Q we know that.
Use Gauss Law , closed ball case it's easy for it to use.
so
∮E dA = Q / ε E is a distance variable , we can take it out
enclose because every point on the ball surface it's same
distance from it's original.
- > E A = Q / ε and Q = ∫∫∫ρdV = ∫∫∫ar r^2 sinδdrdθdδ
enc enc= 2π*2*r^4*a /4
surface of ball = 4πr^2
πr^4 a Q r^2 Qenc r^4
- > E = --------- = ----------- (------ = ------ )
4επr^2 4πεR^4 Q R^4
and by using same way to integral whole ball Q = πR^4a
It's obvious to see what mistake you have done.
You say Q / whole ball = q / 4/3πr^3 it's wrong.
this means all of point one the ball have same ρ density
but this case ρ= A r not a constant..
so you can't use the equation.
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