proton with the mass m is projected into a uniform electric field that points ve

Question

proton with the mass m is projected into a uniform electric field that points vertically upward and has magnitude E. The initial velocity of the proton has a magnitude v0 and is directed at an angle ? below the horizontal.

a)

Find the maximum distance hmax that the proton descends vertically below its initial elevation. You can ignore gravitational forces.

Express your answer in terms of the variables m, ?, v0, E, e.

b)

After what horizontal distance d does the proton return to its original elevation?

Express your answer in terms of the variables m, ?, v0, E, e.

c) Find the numerical value of hmax if E = 500N/C , v0 = 4.40

Explanation / Answer

a) F = q*E

m*a = q*E

a = q*E/m

we know, Hmax = vo^2*sin(theta)/(2*a)

= (m/2*q*E)*vo^2*sin(theta)

b) Range,d = vo^2*sin(2*theta)/a

= (m/q*E)*vo^2*sin(2*theta)

c) Hmax = (1.67*10^-27/1.6*10^-19*500)*(4.4*10^4)^2*sin(30)

= 0.02 m or 2 cm

d) d = (1.67*10^-27/1.6*10^-19*500)*(4.4*10^4)^2*sin(2*30)

= 0.035 m or 3.5 cm

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