A proton is fired directly into a uniform electric field that is able to bring t

Question

A proton is fired directly into a uniform electric field that is able to bring the charge momentarily to rest. (a) What must be the orientation of the electric field relative to the velocity of the proton? (b) What type of charge configuration would generate a uniform electric field? (c) If the proton is initially traveling at 5% the speed of light, what field strength will stop the charge within 1.00 cm of entering the field region? (d) How much work does the electric field do in stopping the proton?

Explanation / Answer

1) The electric field should be in opposite direction to the velocity of proton

2) A infinitely large plane of uniformly distributed charges will give a uniform electric field

3) 5% of speed of light = 0.05*3*10^8 = 1.5*10^7 m/s

KE of proton = 0.5*m*v^2 = 0.5*1.67*10^-27 * (1.5*10^7)2 = 1.8787*10^-13 J

Force on proton upon entering field = qE where E is electric field q = 1.6*10^-19 C

So if proton has to stop at 1cm

1.6*10^-19 * E * 0.01 = 1.8787*10^-13 J

So E should be greater than 1.174*10^8 N/C to stop proton within 1cm

4) Work done = loss in KE = 1.8787*10^-13 J

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