Answer:
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 =

=

= 357.1 V
c. magnitude =

= 4 463 N/C
d. zero