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I asked this question because I think that acceleration could be positive (though I may be wrong.)

The electric field near the Earth's surface has magnitude of about [itex]150 \text{N/C} [/itex]. What is the acceleration of the proton?

Clearly, we need to use these formulas

[itex]E = \dfrac{F}{q}[/itex]

[itex]F = ma[/itex]

Combine the first two equations, we obtain

[itex]E = \dfrac{ma}{q}[/itex]

Let [itex]m[/itex] be the mass of the proton and [itex]q[/itex] be the charge of the proton. Then,

[itex]a = \dfrac{Eq}{m}[/itex]

Thus by substitution, I obtain

[itex]a = \dfrac{(150 \mbox{N/C})(-1.6 \times 10^{-19} \mbox{C})}{1.67 \times 10^{-27} \mbox{kg}}[/itex]

[itex]a \approx -1.44 \times 10^{10} \mbox{m}/\mbox{s}^2[/itex]

## Homework Statement

The electric field near the Earth's surface has magnitude of about [itex]150 \text{N/C} [/itex]. What is the acceleration of the proton?

## Homework Equations

Clearly, we need to use these formulas

[itex]E = \dfrac{F}{q}[/itex]

[itex]F = ma[/itex]

## The Attempt at a Solution

Combine the first two equations, we obtain

[itex]E = \dfrac{ma}{q}[/itex]

Let [itex]m[/itex] be the mass of the proton and [itex]q[/itex] be the charge of the proton. Then,

[itex]a = \dfrac{Eq}{m}[/itex]

Thus by substitution, I obtain

[itex]a = \dfrac{(150 \mbox{N/C})(-1.6 \times 10^{-19} \mbox{C})}{1.67 \times 10^{-27} \mbox{kg}}[/itex]

[itex]a \approx -1.44 \times 10^{10} \mbox{m}/\mbox{s}^2[/itex]

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