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The electric field between two parallel plates has a magnitude of 180 N/C. The two plates are 2.5 cm apart. Recall that the magnitude of the charge on an electron is 1.602 × 10-19 C.
How much work is done to move an electron from the positive plate to the negative plate?

4.0 × 10-21 J
7.2 × 10-19 J
4.5 J
7200 J

Sagot :

Answer:

b

Explanation:

batolisis

The amount of work done to move an electron from the positive plate to the negative plate is :  7.2 * [tex]10^{-19}[/tex] J

Step 1: Given data

magnitude of two parallel plate ( E ) = 180 N/C

Distance between plates ( d ) = 2.5 cm

magnitude of charge on electron  ( e ) = [tex]1.602 * 10^{-19}[/tex]  C

Step 2 :  determine the work done

Work done = qV = еEd

                           = [tex]1.602 * 10^{-19}[/tex]  *  180 * 2.5

                           = 720 * [tex]10^{-22}[/tex] J  ≈   7.2 * [tex]10^{-19}[/tex] J

Hence the amount of work done to move an electron from the positive plate to the negative plate  =  7.2 * [tex]10^{-19}[/tex] J

learn more about electric field : https://brainly.com/question/11509296

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