Question

圖為電場(chǎng)的電場(chǎng)線，現(xiàn)將一電量q=-3.0×10^-9C的負(fù)電荷從A點(diǎn)移到B點(diǎn)、從B點(diǎn)移到C點(diǎn)，從C點(diǎn)移到D點(diǎn)電場(chǎng)力做功分別為：W_AB=3.0×10^-8J、W_BC=1.5×10^-8J、W_CD=9.0×10^-9J．若取C點(diǎn)電勢(shì)為零，試求A、B、D三點(diǎn)電勢(shì)．

Answer 1

AB間的電勢(shì)差為U_AB=

WAB

q

=

3×10-8

-3×10-9

V=-10V
BC間的電勢(shì)差U_BC=

WBC

q

=-5V，CD間的電勢(shì)差U_CD=-3V
則AC間電勢(shì)差為U_AC=U_AB+U_BC=-15V，DC間電勢(shì)差為U_DC=-U_CD=3V
故A、B、D三點(diǎn)電勢(shì)分別為-15V，-5V和3V．
答：A、B、D三點(diǎn)電勢(shì)分別為-15V，-5V和3V．