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Classical Mechanics Level 2

A 9-V battery has an electric potential difference of 9 V 9\text{ V} between the positive and negative terminals. How much kinetic energy in J would an electron gain if it moved from the negative terminal to the positive one?

Details and Assumptions:

  • The charge on the electron is − 1.6 × 1 0 − 19 C -1.6 \times 10^{-19}~\mbox{C} .
  • You may assume energy is conserved (so no drag or energy loss due to resistance for the electron).


The answer is 1.44E-18.

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David Mattingly by David Mattingly

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5 solutions

Fajrul Falah
Aug 9, 2013

V=W/q

W=V.q

W = 9 x 1,6 x 10^-19

W = 1,44 x 10^-18

4 Helpful 1 Interesting 0 Brilliant 0 Confused
Chin Fong Wong
Aug 5, 2013

By Work-Energy Theorem, E = W = VQ = (9 x 1.6e-19) = 1.44e-18 J

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Jiunn Shan Tan
Aug 4, 2013

E = QV

E= (1.6E-19)(9)

E = 1.44E-18

1 Helpful 0 Interesting 0 Brilliant 1 Confused
Jerby Cristo
Aug 10, 2013

The amount of kinetic energy in J can be computed by multiplying the potential difference and the charge of the electron, (9volt X 1.6E-19 = 14.4E-19 or 1.44E-18 Joules of energy)

0 Helpful 0 Interesting 0 Brilliant 1 Confused
Ahmed Lo'ay
Aug 5, 2013

E = Q*V

E = 1.6E-19 * 9 = 1.44E-18 J

0 Helpful 0 Interesting 0 Brilliant 1 Confused

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