=dfrac (Q)(4pi {varepsilon )_(0)r}-|||-(D) =dfrac (Q)(4pi {varepsilon )_(0)(r)^2
dfrac (q)(4pi varepsilon oR)-|||-C. dfrac (q)(4pi {varepsilon )_(0)(R)^2}-|||-D.
C-|||-R-|||-+Q-|||-(B) dfrac (qQ)(4pi {varepsilon )_(0)R} , -dfrac (qQ)(4pi {var
半径为r的均匀带电球面1,带电量为q;其外有一同心的半径为R的均匀带电球面2,带电量为Q,则此两球面之间的电势差U1−U2为( ) A.q4πε0(1
2球形电容器两同心金属球面的半径分别为R1,R2,各自带有等量异号电荷 +a-0, 置于真空当中,则[]-|||-R1-|||-R 2 +Q -Q-|||--|
dfrac (qQ)(4pi {varepsilon )_(0)(d)_(1)} B. dfrac (qQ)(2pi {varepsilon )_(0)(d)_
设无穷远处电势为零,则球壳内-|||-各点的电势U可表示为:( =dfrac (1)(4pi {varepsilon )_(0)}-|||-(A) lt -Kd
dfrac (1)(2)pi .-|||-C.π-|||-D. dfrac (5)(4)pi .-|||-0
置于真空当中,则-|||-该电容器的电容为 pi (varepsilon )_(0)cdot (dfrac ({R)_(1)(R)_(2)}({R)_(2)-(
两个同心球面的半径分别为R1和R2,各自带有电荷Q1和Q2.求:(1)各区域电势分布.(2)两球面间的电势差为多少?两个同心球面的半径分别为R1和R2,各自带有