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If 60 V is applied to an electric field which moves a charge with an energy of 4.03 x 10 -16 J, determine the magnitude of the charge in the field.
in Physics by Diamond (74,874 points)
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4 Answers

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V= E/Q

Q= E/V

=4.03*10^-16/60

=6.7166667E-18 C
by Wooden (2,658 points)
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V = 60 V Q = 4,03X10^-16 J E? V = E/Q 60 = E/ 4,03 x 10^-16 E =2,418 x10^-14 C
by Wooden (516 points)
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V = E/Q Q=E/V =4,03*10^-6/60 =6,71666666/*10^-8 C
by Wooden (622 points)
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V=E/q

q=E/V

= 4.03 x 10 -16 J/60V

= 6.72 x 10^-18C
by Wooden (1,084 points)

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