Potential difference to bring a proton to rest

AI Thread Summary
To determine the potential difference required to reduce a proton's speed from 3.5×10^5 m/s to half that speed, the conservation of energy principle is applied. The initial kinetic energy (KE) of the proton is calculated using the formula KE = 0.5 * m * v^2. The electric potential energy (EPE) is represented by EPE = q * V, where q is the charge of the proton. Initial attempts to calculate the potential difference resulted in incorrect values, but after reevaluating the energy conservation equation, the correct potential difference was found. The discussion emphasizes the importance of correctly applying conservation of energy in solving such problems.
foreignfishes
Messages
3
Reaction score
0

Homework Statement


A proton has an initial speed of 3.5×10^5 . What potential difference is required to reduce the initial speed of the proton by a factor of 2?


Homework Equations


KE=.5*m*v^2
EPE=q*V


The Attempt at a Solution


I did KE=.5(1.67x10^-27)(3.5x10^5/2)^2 and then divided that by the charge of a proton to get V, and I got 160 V which is wrong. Then I tried squaring the velocity first, then dividing it by 2 and I got 320 V, which is also not right. I'm stuck.
 
Physics news on Phys.org
this is just conservation of energy

electric potential starts at zero plus kinetic energy starts at some KEi

this must be equal to some electric potential energy plus half of KEi
 
SHISHKABOB said:
this is just conservation of energy

electric potential starts at zero plus kinetic energy starts at some KEi

this must be equal to some electric potential energy plus half of KEi

Thank you so much! I didn't even think of that. I got the answer.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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