PHYSIC 難題40

Potential V1 at point A is given by,
V1 = [k(3x10^-6)/0.5 + k(-2x10^-6)/0.5] volts
where k is the electrostatic constant (= 9x10^9 N.m^2/C^2)
hence, V1 = [(10^-6)k/0.5].(3-2) volts = (2x10^-6)k volts

Potential V2 at point B is,
V2 = [k(3x10^-6)/1 + k(-2x10^-6)/1] volts
i.e. V2 = (10^-6)k volts

Change in potential when the -4uC charge moves from A to B
= V2 - V1
= [(10^-6)k - (2x10^-6)k] volts
= -(10^-6)k volts

Energy required = (-4x10^-6) x [ -(10^-6)k] J
= (4x10^-12) x (9x10^9) J
= 0.036 J