静止电荷的电场作业题目及解答.ppt

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1、781013172024第七章静电场作业题及解答(作业题个数13)263948525764作业题总目录Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.7-7如图所示的电荷分布称为电四极子，它由两个相同的电偶极子组成。证明在电四极子轴线的延长线上离中心为r(r＞＞re)的P点处的电场强度为式中Q=2qre2称为这种电荷分布的电四极矩。π4rQ0e34E=P-q+2qrere-qr题号结束Evaluationonly.CreatedwithAs

2、pose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.已知：r<

3、le5.2.0.0.Copyright2004-2011AsposePtyLtd.3Q==qπε40rer426πε40r4reE=qπε20rer6r23.题号结束Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.7-8长l=15cm的直导线AB上均匀地分布着线密度为l=5×10-9C/m：的电荷（如图）。求：（1）在导线的延长线上与导线一端B相距d=5cm处P点的场强；（2）在导线的垂直平分线上与导线中点相距d=5cm处Q点的场强。..Q

4、PBAddl题号结束Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.1xlπx2ε40dEd=xlπx2ε40dE=òdd+l()lπε40=dd+l11=()×1029×109×5.0×10-95.02011=6.75×102V/m已知：l=15cm，l=5×10-9C/m,d=5cm求：EP解：(1).PBAdlxxdx题号结束Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5Client

5、Profile5.2.0.0.Copyright2004-2011AsposePtyLtd.1qxlcosπr2ε40=dEdEy=drd12==+()qcosdx22d.1xlπε40dEd=+()dx2212+()dx22d..1xlπε40d=32+()dx22d(2)由对称性Ex=0.QdlEdxdr题号结束Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Ed.1xlπε40d=32+()dx22dlπε40E=ò--l/2dxd3

6、2+()dx22l/2lπε40=dl12+()ld2221.20.155.0×10-2+=()()5.0×10-9×0.15×9×1095.0×10-22212=1.50×103V/m题号结束Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.题号结束ayxoq7-10用细绝缘线弯成的半圆环，半径为a其上均匀带电正电q。试求：圆心O点处的电场强度。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.

7、5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.题号结束Ey=0òEx=dEqsin=òEda=dlqddEdaεπ40=q2a=lqdaεπ40=2alqdπ0qsin=òaεπ40lqdπ0qsin=òaεπ40lqdqcos=aεπ40lπ0=aεπ20lπ=aε2022qql=aπdq=ldl解：由对称性aEdqdqyxoEvaluationo