A particle with a charge of –1.2 nC (nano-coulomb, or 10-9C) is located at a distance of 2.3 cm from a particle with charge of 2.8 nC (2.8 × 10-9 C). The first particle is released from rest. The second particle is fixed at its location. What is the change in electric potential energy when the first particle has moved 1.0 cm?

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whitneytr12 whitneytr12 · Answer 1 · 2020-10-26T23:22:29+01:00

Answer:

So the change will be express as a difference of both electric potential:

[tex]\Delta U = U_{2}-U_{1}=1.011*10^{-6} J[/tex]

Explanation:

The electric potential is given by:

[tex]U=k\frac{q_{1}q_{2}}{r}[/tex]

The electric potential at a distance of 2.3 cm will be:

[tex]U_{1}=9*10^{9}\frac{1.2*10^{-9}2.8*10^{-9}}{0.023}[/tex]

[tex]U_{1}=1.315*10^{-6}J[/tex]

Now, the electric potential when the first particle has moved 1.0 cm is.

Here, the distance between the two particles will be 2.3 cm - 1 cm = 1.3 cm

[tex]U_{2}=9*10^{9}\frac{1.2*10^{-9}2.8*10^{-9}}{0.013}[/tex]

[tex]U_{2}=2.326*10^{-6} J[/tex]

So the change will be express as a difference of both electric potential:

[tex]\Delta U = U_{2}-U_{1}=1.011*10^{-6} J[/tex]

I hope it helps you!