Question

Question #8e4e9

Answer 1

`1.02xx10^8" V"`, rounded to two decimal places

Explanation:

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Let the voltage required between the horizontal plates be `=V`
Electric field between the plates`=V/d=V/0.1=10V " Vm"^-1`

Force due to electric field on the proton`=qvecE`
`=1.6 xx 10^-19xx10V`
`=1.6 xx 10^-18V" N"`

Downwards force due to weight of proton`=mg`
`1.67 xx10^-27xx9.81=1.638xx10^-26" N"`

To hold the proton motionless both forces must be balanced. We get
`1.6 xx 10^-18V=1.638xx10^-26`
`=>V=(1.638xx10^-26)/(1.6 xx 10^-18)`
`=>V=1.02xx10^8" V"`, rounded to two decimal places