(a) [tex]1.50\cdot 10^{-6}J[/tex]
According to the work-energy theorem, the work done by the electric force is equal to the kinetic energy gained by the particle, so we can write:
[tex]W=K_f - K_i[/tex]
where
W is the work done by the electric force
[tex]K_f[/tex] is the final kinetic energy of the particle
[tex]K_i[/tex] is the initial kinetic energy of the particle
Since the particle starts from rest, [tex]K_i = 0[/tex]. Moreover,
[tex]K_f = 1.50\cdot 10^{-6}J[/tex]
Therefore, the work done by the electric force is
[tex]W=K_f = 1.50\cdot 10^{-6}J[/tex]
(b) -333.3 V
According to the law of conservation of energy, the gain in kinetic energy of the particle must correspond to a loss in electric potential energy, so we can write:
[tex]\Delta K = -\Delta U[/tex]
Where
[tex]\Delta K = 1.50\cdot 10^{-6} J[/tex] is the gain in kinetic energy
[tex]\Delta U[/tex] is the loss in electric potential energy
So we have
[tex]\Delta U = - \Delta K = -1.50\cdot 10^{-6}J[/tex]
The loss in electric potential energy can be rewritten as
[tex]\Delta U = q \Delta V[/tex]
where
[tex]q=+4.50 nC = 4.5\cdot 10^{-9} C[/tex] is the charge of the particle
[tex]\Delta V[/tex] is the change in electric potential over the distance the charge has moved
Solving for [tex]\Delta V[/tex],
[tex]\Delta V= \frac{\Delta U}{q}=\frac{-1.50\cdot 10^{-6}}{4.5\cdot 10^{-9}}=-333.3 V[/tex]
(c) 4166 V/m
The magnitude of the electric field is given by
[tex]E=\frac{|\Delta V|}{d}[/tex]
where
[tex]|\Delta V|[/tex] is the magnitude of the change in electric potential
d is the distance through which the charge has moved
Since we have
[tex]|\Delta V|=333.3 V\\d = 8.00 cm = 0.08 m[/tex]
The magnitude of the electric field is
[tex]E=\frac{333.3}{0.08}=4166 V/m[/tex]
(d) [tex]-1.50\cdot 10^{-6}J[/tex]
The change in electric potential energy of the charge has already been calculated in part (b), and it is
[tex]\Delta U = -1.50\cdot 10^{-6}J[/tex]
