An electric field of strength E = 5800 N/C is directed along the + x -axis as sh

Question

An electric field of strength E = 5800 N/C is directed along the +x-axis as shown.
An electron is initially at rest at point B.

1)What is the magnitude of the force on the electron due to the electric field?

2)What is the work done by the electric field on the electron after it has moved a distance of 4cm from point B to point A?

3)What is the kinetic energy of the electron at point A?

4)What is the kinetic energy of the electron at this instant measured in eV?

Can you also show how to do each one?

Explanation / Answer

force = q E = 1.6 x 10^-19 x 5800 = 9.28 x 10^-16 N

work = force x distance = 9.28 x 106-16 N x .04 m = 3.712 x 10^-17 Nm

work done = KE

the work done on the particle increases its kinetic energy

thus KE = 3.712 x 10^-17 J

1 J = 6.24 x 10^18 eV

KE in eV = 231.63 eV

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