NasuSama
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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 150 \text{N/C}. What is the acceleration of the proton?
Clearly, we need to use these formulas
E = \dfrac{F}{q}
F = ma
Combine the first two equations, we obtain
E = \dfrac{ma}{q}
Let m be the mass of the proton and q be the charge of the proton. Then,
a = \dfrac{Eq}{m}
Thus by substitution, I obtain
a = \dfrac{(150 \mbox{N/C})(-1.6 \times 10^{-19} \mbox{C})}{1.67 \times 10^{-27} \mbox{kg}}
a \approx -1.44 \times 10^{10} \mbox{m}/\mbox{s}^2
Homework Statement
The electric field near the Earth's surface has magnitude of about 150 \text{N/C}. What is the acceleration of the proton?
Homework Equations
Clearly, we need to use these formulas
E = \dfrac{F}{q}
F = ma
The Attempt at a Solution
Combine the first two equations, we obtain
E = \dfrac{ma}{q}
Let m be the mass of the proton and q be the charge of the proton. Then,
a = \dfrac{Eq}{m}
Thus by substitution, I obtain
a = \dfrac{(150 \mbox{N/C})(-1.6 \times 10^{-19} \mbox{C})}{1.67 \times 10^{-27} \mbox{kg}}
a \approx -1.44 \times 10^{10} \mbox{m}/\mbox{s}^2
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