$5\mathrm{.}5.$ A circular plate has a radius of $11$ cm and a total charge of $28$ C$.$ A dust particle of charge $4$ C is initially at rest, a distance of $5$ mm from the center of the plate. The positively charged plate repels the positively charged particle and after a short time, the particle's distance from the plate has tripled.

What is the voltage difference between the particle's initial and final positions? $[$Pay attention to the sign of this difference!$]$

Yes, Computer gets: $-416\mathrm{.}07$ kV

Hint: First use the information about the total charge and radius of the plate to calculate the plate's surface charge density. Use the formula for the voltage produced by the plate to calculate the voltage difference between the plate and the particle's initial position. Do the same to compute the voltage difference between the plate and the particle's second position. Then compute the difference. Is the sign what you would expect?

$6\mathrm{.}6.$ The particle's mass is $15$ mg$.$ How fast is the particle moving when it reaches the second position? $[1$ mg $=106106$ kg$.]$