Question

A 2 V battery is connected to two parallel plates a distance 55 cm apart. If a `-4.30 xx 10^-4 C` charge was placed in the electric field, what force would it feel?

What work is needed to move the `-4.30 xx 10^-4 C` charge from 0.138 m from the ground plate to 0.413 m from the ground plate?

Answer 1

Electric field is given by

`vecE=V/dhat r`
direction of `hat r` is from positive plate to negative plate.

Inserting given values we get

`|vecE|=2/0.55Vm^-1` .....(1)

Force experienced by a charged particle `q` in an electric field is

`vecF=qvecE`

Hence , using (1) we get

`vecF=2/0.55xx(-4.30xx1^-4)hat r`
`|vecF|=1.56xx10^-3N`
Direction is opposite to that of `hat r`, i.e., towards positive plate

Work done in moving the charge is

`W=vecFcdot vecs`

As the angle between displacement and force vectors is `0^@`,
`costheta` in the dot product `=cos0^@=1`

`:.W=|vecF| |vecs|`
`=>W=1.56xx10^-3xx (0.413-0.138)`
`=>W=1.56xx10^-3xx (0.413-0.138)`
`=>W=4.3xx10^-4J`