Can the acceleration of the proton be negative for this situation?

[NasuSama]

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

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]

 

[mfb]

A proton has a positive charge (and you are off by 10 orders of magnitude with its absolute value).
The magnitude of the acceleration is always positive, the sign of the acceleration depends on the (arbitrary) direction of your coordinate.

 

[NasuSama]

A proton has a positive charge (and you are off by 10 orders of magnitude with its absolute value).
The magnitude of the acceleration is always positive, the sign of the acceleration depends on the (arbitrary) direction of your coordinate.

I get it by now. What about the acceleration of the electron? I calculated it for my online HW, and I found that the acceleration of the electron in the same electric field is [itex]2.6 \times 10^{13} \mbox{m}/\mbox{s}^2[/itex]. The online HW system marks it correct. Is it true that acceleration is also positive?

 

[mfb]

The sign of the acceleration is meaningless if you do not use a specific coordinate system. The acceleration of electrons and protons happens in the opposite direction.

 

[NasuSama]

I see by now. Thanks for your help.