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/MD. What was the change in potential energy of the charge? in J

Answer 1

a. 1.50×10⁻⁶ J

b. 357.1 V

c. 4 463 N/C

d. zero

Step-by-step explanation:

We are given:

q= +4.20 nC

= + 4.20 × 10 ⁻⁹ C

The charge moves from rest to a point in the negative direction. Therefore, the kinetic energy of the charge at the point of origin is Ka = 0

the distance, l = 8 cm

= 0.08 m

a. Work done by electric field = Ka

= 1.50×10⁻⁶ J

b. the change in electric potential =
`(V)/(q)`

=
`(1.50 x 10-6)/(4.20x10-9)`

= 357.1 V

c. magnitude =
`(357.1)/(0.08)`

= 4 463 N/C

d. zero