Question

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

Answer 1

Answer:

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