Place two charged insulating plates parallel to each other. The charge densities on the left and right insulating plates are $$sigma $1=+1\mathrm{.}0$sigma and $$sigma $2=1\mathrm{.}0$sigma $,$ where $$sigma $=1\mathrm{.}6$ nC$/$m$2.$ The vacuum permittivity is $0=8\mathrm{.}85421012$ C$2/($N$$m$2).$

a$)$ Calculate the electric field in region B$.$ The positive direction is from left to right.

EB$=$

The plates are separated by a distance d$=2\mathrm{.}5$ cm$.$ A positively charged particle $($charge q$=1\mathrm{.}1$mu C$),$ with a mass m$=0\mathrm{.}13$ g$,$ starts moving from rest at time t$=0$ from the positively charged plate between the plates.

b$)$ What is the particle's acceleration between the plates?

a$=$

c$)$ At what time does the particle collide with the negatively charged plate?

t$=$

d$)$ What is the particle's collision velocity?

v$=$

Transcribed Image Text:

d + I C | I Я + + + + + A 02 01 d + I C | I Я + + + + + A 02 01

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a The electric field E between two charged parallel plates with charge densities 1 and 2 can be calculated using the formula E 1 2 2 0 Where 0 is the ... View the full answer