大物課件及課后習(xí)題答案8第八章

1、r dA f br dv rQ11rrarr00q 1 ab1rr (dr Af rr dA f br dv rQ11rrarr00q 1 ab1rr (dr Af rrra00aba pr vvr0rnvE )2q q 20ip ( pr vvr0rnvE )2q q 20ip (1iO3E11q i0L()4p2(L)Fdrp(L)qEFdr Aq0vvrrqE00p (L p (L L2Fdr Aq0vvrrqE00p (L p (L L2Pvvrr1qEdrqEdr00p (L p (L r E=brrdl WaWb0f a若b點(diǎn)為勢(shì)能零Wb rvv dlf dlfWa點(diǎn)電勢(shì)能aaq在

2、a點(diǎn)的電勢(shì)能等于將q從點(diǎn)電荷Q的靜電場(chǎng)ar =brrdl WaWb0f a若b點(diǎn)為勢(shì)能零Wb rvv dlf dlfWa點(diǎn)電勢(shì)能aaq在a點(diǎn)的電勢(shì)能等于將q從點(diǎn)電荷Q的靜電場(chǎng)ar 1 bA rf q從b電場(chǎng)力所做的 rb 0aa1W點(diǎn)電荷Q的靜電場(chǎng)中各點(diǎn)的電勢(shì)能r0勢(shì)能零 Eaa2)勢(shì)能零 Eaa2)？能否能用電勢(shì)能Wa 來(lái)表示a點(diǎn)電場(chǎng)的性質(zhì)呢：J（焦耳eV（電子伏Q電aWqaq變a 但aq W勢(shì)能零a點(diǎn)電1)定EaqaQ電aWqaq變a 但aq W勢(shì)能零a點(diǎn)電1)定Eaqa點(diǎn)的Vaa=V伏特2)9勢(shì)能Eaa勢(shì)能Eaab電勢(shì)差b b qEf Eb電勢(shì)差b b qEf Eaaqa與試驗(yàn)電荷無(wú)關(guān)電

3、場(chǎng)在abba bE dlaWaWbb E da a,b將a,ba,a,b將a,ba,bbAq(a b)qEa電勢(shì)疊加原理 電勢(shì)計(jì)PE Qr EP r電勢(shì)疊加原理 電勢(shì)計(jì)PE Qr EP r E dlr22r00rrdl 4 01)2)000rrrr 1)2)000rrrr PPPii dQ i0 EaaQP Rr即rRrQrr v E EQP Rr即rRrQrr v E E rPP0rRr E r E r0QR 4 rP0rR0Rr即 QR 4 rP0rR0Rr即 rr2 rP0r0r的r0R例題：平行板電容器兩板間的電勢(shì)E 例題：平行板電容器兩板間的電勢(shì)E rdEEd E, EE 位長(zhǎng)度電量

4、分別為12，設(shè)外筒電勢(shì)為0位長(zhǎng)度電量分別為12，設(shè)外筒電勢(shì)為021當(dāng)rR 2 2120 12E3rrrrr E dl E 2E1rRPR1R2 rR 0 rrrrr E dl E 2E1rRPR1R2 rR 0 ln212當(dāng)R1 rR2時(shí)rrRrrE2p 0R1rr3 vE當(dāng) p1 1ln rrr3 vE當(dāng) p1 1ln rrR2r筒間即r=R1ln PPRr 01例題：求均勻帶電圓環(huán)軸線上p點(diǎn)的y解方法一dqr電荷元 dq在p例題：求均勻帶電圓環(huán)軸線上p點(diǎn)的y解方法一dqr電荷元 dq在pxodU pRU dU QQQ方法二 EdlUP1E p3(R2 x2)p0rrE dlUP r1E dl1E p3(R2 x2)p0rrE dlUP r1E d