What is a "g" for electric fields?

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A "g" is a unit of acceleration, specifically the acceleration experienced by a mass in free fall near Earth's surface, equivalent to 9.80665 m/s². The discussion clarifies that "g" is not a unit of force but rather measures the acceleration a body feels when subjected to forces. The context involves calculating the electric field strength at a point where a proton experiences an acceleration of 1 million "g's." Participants emphasize the distinction between force and acceleration in this context. Understanding "g" is crucial for accurately interpreting concepts related to electric fields and forces.
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What is a "g"?

What is the electric field strength at a paoint in space where a protone (m=1.67x10^-27 kg) experiences an acceleration of 1 million "g's"?

My only question is:
What is a "g"?
 
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Its a unit of force equal to the force exerted by gravity. Its basically the amount of force a body feels when its accelerated.
 
ranger said:
Its a unit of force equal to the force exerted by gravity. Its basically the amount of force a body feels when its accelerated.

No, it is a unit of acceleration, not a unit of force.

It is the acceleration of a mass falling freely near the Earth's surface (with no air resistance, etc).

1g = 9.80665 m/s^2.
 
Thanks guys. I can't believe I didn't know that "g" meant gravity.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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