Answer
(a) negative x direction
(b) $3.8cm/s$
(c) less than
Work Step by Step
(a) As the particle has a positive charge, it will move in the direction of the electric field that is in the negative x direction.
(b) We can find the required speed as
$\frac{1}{2}mv^2=-qE\Delta x$
This simplifies to:
$v=\sqrt{\frac{-2qE\Delta x}{m}}$
We plug in the known values to obtain:
$v=\sqrt{\frac{-2(0.045\times 10^{-6})(1200)(-0.050)}{0.0038}}$
$v=3.8cm/s$
(c) Since the speed is directly proportional to the square root of the distance, the increase in speed over the second 5.0 cm is less than that of the first 5.0cm.
