A Proton Between Two Parallel Plates

Electricity and Magnetism Level 2

A proton is placed midway between two parallel metallic plates which are $0.2$ meter apart. The plates are connected to an $80$ volt battery. What is the magnitude of the electric force on the proton,in newtons?

Note: A proton has an electric charge of $1.6\times10^{-19}$ Coulomb.

3.2\times10^{-16}

3.2\times10^{-20}

6.4\times10^{-17}

1.28\times10^{-16}

1 solution

Fahim Shahriar Shakkhor
Sep 23, 2014

The plates were charged oppositely by the battery and this produce a uniform electric field between the plates. Its intensity is equal to the difference of potential between the plates to the distance between the plates :

$E = \dfrac{V}{d} = \dfrac{80V}{0.2m} = 400 N/C$

The field intensity is defined as the force on a unit positive charge placed in the field. So,

$E = \dfrac{F}{q} \\ \Rightarrow F = qE = (1.6 \times 10^{-19} C) \times 400 N/C = \boxed{6.4 \times 10^{-17}N}$

Why d=0.2m not 0.1 meter. As equally means on the midway point so \frac{0.2}{2]=0.1

Nafees Zakir - 6 years, 8 months ago

@Nafees Zakir i agree with you @Fahim Shahriar Shakkhor can you explain why this is?

Mardokay Mosazghi - 6 years, 7 months ago

Is there anything wrong I can't see LaTex written by me but can see written by others.

Nafees Zakir - 6 years, 8 months ago

A Proton Between Two Parallel Plates

1 solution

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