Answer:
1.5 x 10⁻⁶ J
Explanation:
q = magnitude of charge on the charge = 1 x 10⁻⁶ C
ΔV = Potential difference between positive and negative terminals of battery = 1.5 Volts
W = Amount of work done to move the charge from negative terminal to the positive terminal
Work done is given as
W = q ΔV
W = (1 x 10⁻⁶) (1.5)
W = 1.5 x 10⁻⁶ J
