# No clear on concept

not clear on concept

Q Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1650 km/s, measured relative to the earth. find the maximum electrical force that these protons will exert on each other?
K_a + -U_a= K_b + U_b and K_a=0, U_b=0
-U_a=K_b
K_b= 1/2mv^2
U_a= (1/(4*pi*epsilon_0)((q*q_0)/r)
F=qE
is the radius of the earth r and how do i get the force from potential energy?

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Doc Al
Mentor
gmuniz said:
K_a + -U_a= K_b + U_b and K_a=0, U_b=0
If this is meant to be conservation of energy, rewrite it like this:
$${KE}_i + U_i = {KE}_f + U_f$$
where K_i is the initial KE; K_f = 0; U_i = 0; U_f is the electrical potential energy when the protons have momentarily stopped.

Use this to find the distance of closest approach, where KE = 0. Once you have that distance, use Coulomb's law to find the force.

is r=((q^2/4*pi*epsilon_0)(1/K_i)

Doc Al
Mentor
No. Solve for r by equating the intial KE (of both protons) with the final electric potential energy, which depends on r.