Question

在某電場(chǎng)中將電量為-3×10^-6c的點(diǎn)電荷從A移到B，電場(chǎng)力做功6×10^-4J；從B移到C，電場(chǎng)力做功一5.4×lO^-4J，則此負(fù)電荷在這三點(diǎn)中的    點(diǎn)時(shí)電勢(shì)能最低，A、C兩點(diǎn)間的電勢(shì)差U_AC=    V．

Answer 1

【答案】分析：根據(jù)電勢(shì)差公式求出任意兩點(diǎn)間的電勢(shì)差，并判斷兩點(diǎn)間電勢(shì)的高低．A、C兩點(diǎn)間的電勢(shì)差U_AC=U_AB+U_BC．
解答：解：在電場(chǎng)中將電量為-3×10^-6C的點(diǎn)電荷從A移到B，電場(chǎng)力做功6×10^-4J，則A、B間電勢(shì)差U_AB==-200V，說(shuō)明A點(diǎn)電勢(shì)低于B點(diǎn)電勢(shì)．從B移到C，電場(chǎng)力做功-5.4×lO^-4J，則B、C間電勢(shì)差U_BC==180V，說(shuō)明B點(diǎn)電勢(shì)高于C點(diǎn)電勢(shì)．A、C兩點(diǎn)間的電勢(shì)差U_AC=U_AB+U_BC=-20V，說(shuō)明A點(diǎn)電勢(shì)低于C點(diǎn)電勢(shì)．所以在這三點(diǎn)中的B點(diǎn)電勢(shì)最高，負(fù)電荷在B點(diǎn)電勢(shì)能最低．
故答案為：B，-20
點(diǎn)評(píng)：對(duì)于電勢(shì)高低的比較，可以根據(jù)電勢(shì)差的正負(fù)比較電勢(shì)的高低，也可以根據(jù)電場(chǎng)力做功的正功與電荷的電性判斷兩點(diǎn)間電勢(shì)的高低．