A particle with a charge of +4.20 nC is in a uniform electric field E⃗ directed to the negative x direction. It is released from rest, and after it has moved 8.00 cm , its kinetic energy is found to be 1.50×10−6 J .

A. What work was done by the electric force?
B. What was the change in electric potential over the distance that the charge moved?
C. What was the magnitude of E? in V/M
D. What was the change in potential energy of the charge? in J

Explanation:

Charge on the particle

= 4.2 x 10⁻⁹ C

K.E = 1.5 x 10⁻⁶ J

A ) Work done by electric field is converted into kinetic energy

So work done by electric force = 1.5 x 10⁻⁶ J

B ) Kinetic energy attained = potential diff x charge

1.5 x 10⁻⁶ = PD x 4.2 x 10⁻⁹

Potential diff ( PD ) = .357 x 10³

=357 V

C )

Magnitude of electric field

= potential difference / distance ( potential gradient )

= 357 / (8 x 10⁻²)

= 44.625 x 10² V/m

D )

Change in the potential energy of charge

= work done by electric field

= 1.5 x 10⁻⁶ J

what experimental evidence led scientists to change from the previous model to this one?

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the colors of light emitted from heated atoms had very specific energies.

experiments with water vapor showed that elements combine in specific proportions.

cathode rays were bent in the same way whenever a magnet was brought near them.