The charge of the first particle is -1.5 nC and the charge of the second particle is 2.5 nC. The particles are released are initially separated by 2.5 cm, having first particle released from rest position. The second particle is fixed at its location. What is the change in electrical potential energy when the first particle moved 1.0 cm?

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Answer: 1.4 seconds which is the third optionexplanation: current is defined as the flow of charge per timetherefore, the equation that relates current, charge and time can be expressed as follows: i = δq / δtwhere: i is the current given = 1.8 ampereδq is the charge given = 2.5 coulombδt is the time required for the flow of the given charges to give the 1.8 ampere.substitute with the givens in the above equation to get the time as follows: 1.8 = 2.5 / δtδt = 2.5 / 1.8 = 1.38888 which is approximately 1.4 secondshope this was !