Question

將一個(gè)電量-2×10^-8C的點(diǎn)電荷，從零電勢(shì)點(diǎn)O移到M點(diǎn)需克服電場(chǎng)力做功4×10^-8J，則M點(diǎn)電勢(shì)φ_M=

-2v

；若將該電荷從M點(diǎn)再移至N點(diǎn)，電場(chǎng)力做功1.4×10^-7J，則N點(diǎn)電勢(shì)φ_N=

5v

，M、N兩點(diǎn)間的電勢(shì)差U_MN=

-7v

．

Answer 1

分析：電場(chǎng)力做功與電勢(shì)能的關(guān)系，再有電勢(shì)與電勢(shì)能的關(guān)系求某點(diǎn)的電勢(shì)高低，電勢(shì)差與電勢(shì)的關(guān)系．

解答：解：根據(jù)電場(chǎng)力做功與電勢(shì)能的關(guān)系，Em=W_MO
    由電勢(shì)與電勢(shì)能的關(guān)系，則M點(diǎn)的電勢(shì)為     qφ_M=Em，φ_M=

Em

q

=

4×10-8

-2×10-8

=-2V
   同理  W_MN=E_M-E_N   即E_N=W_MN-E_M
        E_N=qφ_N          
    解得φ_N=5V
   電勢(shì)差與電勢(shì)的關(guān)系：U_MN=φ_M-φ_N=-2V-5V=-7V 
故答案為：-2v，5v，-7v

點(diǎn)評(píng)：考查了電場(chǎng)力做功與電勢(shì)能、電勢(shì)、電勢(shì)差的關(guān)系，注意負(fù)電荷的情況．