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A 20 nC charge is moved from a point where V= 130 V to a point where V= -70 V How much work is done by the force that moves the charge?Express your answer with the appropriate units.

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ANSWER

4 μJ

EXPLANATION

Given:

• The charge, q = 20 · 10⁻⁹ C

,

• The intital potential, V₁ = 130 V

,

• The final potential, V₂ = -70 V

Find:

• The work done on the charge, W

The work done is,

[tex]W=q\cdot\Delta V[/tex]

Replace the known values,

[tex]W=20\times10^{-9}C\cdot(-70-130)V=-4\cdot10^{-6}J=-4\mu J[/tex]

The work is negative because the work is not done by the charged particle, but by an external force. Hence, the work done by an external force is 4 μJ.

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