Question

A proton is shot in through the side of a parallel plate capacitor with kinetic energy 8.35x10^(-20) joules. The electric field is providing a force on the proton of 3.2x10^(-16) newtons. The mass of a proton is 1.66x10^-27) kilograms. (a) Determine the magnetic field (strength and direction) in order for the proton to make it through to the other side on a straight-line path. (b) If the region of the magnetic field found in "a" extends past the capacitor, determine the radius of the path of the proton and sketch it on the diagram. (c) Explain what happens to the proton if it is shot in with more kinetic energy while all the other parameters remain the same. (d) Explain what happens to a more massive particle that is shot in with the same speed while all other parameters remain the same.

Answer 1

(A) KE = m v^2 / 2
8.35 x10^-20 = (1.66 x 10^-27) v^2 /2

v = 10^4 m/s

if Fe = Fb and they are acting in opposite direction.

Fb = q v B

3.2 x 10^-16 = (1.6 x 10^-19)(10^4) (B)

B = 0.20 T  

perpendicular to the page.

(b) then Fb = m v^2 / r

(3.2 x 10^-16) = (1.66 x 10^-27)(10^4)^2 / r

r = 0.52 x 10^-3 m OR 0.52 mm

(C) r = m v / q B

is v is greater than then radius for circular path will be greater.

(d) if m is greater then also radius will be greater.