※ 引述《jack8366 (卷毛)》之铭言： : ※ 引述《wyech (wicky)》之铭言： : : 时间: Mon Jul 5 16:49:24 2010 : : 一半径为R的非导电实心球，总电荷为Q，电合密度分布为ρ(r)=Ar C/m3，其中A为常数。 : : 以Q来表示距中心r处的电场当r<R时 : : -- : : ◆ From: 114.41.13.44 : 我想请问一下~!这题答案真的是 Qr^2/4piεR^4 吗? : 因为我翻了原文书和自己算 答案都是 Qr/4piεR^3 耶.... : 有没有同学可以帮我解一下这题...因为没解出来害我实在是睡不着 / _> \ : 到底是我算错了吗.... : 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带入 : => E = Qr/4piεR^3 : ※ 编辑: jack8366 来自: 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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