Question

將一個(gè)電荷量為-2×10^-8C的點(diǎn)電荷，從零電勢(shì)點(diǎn)S到M點(diǎn)要克服電場(chǎng)力做功4×10^-8，則M點(diǎn)電勢(shì)U_M=

-2V

．

Answer 1

分析：根據(jù)移動(dòng)試探電荷電場(chǎng)力做的功，求出S、M兩點(diǎn)間的電勢(shì)差U_SM，根據(jù)U_SM=φ_S-φ_M求解M點(diǎn)電勢(shì)U_M．

解答：解：由題，q=-2×10^-8C，W_SM=-4×10^-8J，則S、M兩點(diǎn)間的電勢(shì)差U_SM=

WSM

q

=

-4×10-8

-2×10-8

V=2V
由于U_SM=φ_S-φ_M，S點(diǎn)電勢(shì)為零，φ_S=0，則M點(diǎn)電勢(shì)φ_M=-U_SM=-2V．
故答案為：-2V

點(diǎn)評(píng)：求解電勢(shì)，常常先求出該點(diǎn)與零電勢(shì)點(diǎn)間的電勢(shì)差，再求電勢(shì)．要注意電勢(shì)差的順序．