The Electric Field Strength of a Proton Accelerated at 1 Million g

AI Thread Summary
The discussion centers on calculating the electric field strength experienced by a proton undergoing an acceleration of 1 million g, equating to 9.8x10^6 m/s^2. Participants clarify that the relationship between force, mass, charge, and electric field strength is essential for this calculation. The formula used is F = ma = qE, leading to the conclusion that the electric field strength can be derived from the mass of the proton, its acceleration, and its charge. The final calculation suggests that the electric field strength is approximately 0.1 N/C. This highlights the importance of correctly applying physical principles to determine electric field strength in high-acceleration scenarios.
joej
Messages
29
Reaction score
0
What is the electric field strength at a point in space where a proton experiences an acceleration of 1 million "g"?

I'm completely lost on this one I know it's bound to be something simple but I just can't figure out where to start with it, well other than that 1 million "g" should be 9.8x10^6 m/s^2

I'm also not quite sure, but I'm guessing this will have something to do with gravitational fields.
 
Physics news on Phys.org
The question seems to infer that the proton is accelerated by the electric field.

Think about the relationship of Force to the electric field strength and as a function of mass and acceleration.

\vec{F} = m\vec{a} = q\vec{E}
 
okay, so I get...

ma = (1.64x10^-21)/(1.6x10^-19) = 0.01N/C

would that be right?
 
joej said:
okay, so I get...

ma = (1.64x10^-21)/(1.6x10^-19) = 0.01N/C

would that be right?

U missed one order of size:
|\vec{E}|=\frac{m_{proton}|\vec{a}|}{q_{proton}}
|\vec{E}|=\frac{1.67\cdot 10^{-27}Kg\cdot 9.8\cdot 10^{6} ms^{-2}}{1.6\cdot 10^{-19}C}\sim 0,1N/C

Daniel.
 
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}...
Back
Top