A pair of oppositely charged parallel plates are separated bya distance of 0.05m

Question

A pair of oppositely charged parallel plates are separated bya distance of 0.05m with a potential difference of 550V between theplates. A proton is released from rest at the positive plate at thesame time that an electron is released from rest at the negativeplate. Disregard any interation between the proton and theelectron. a) How long does it take for the paths of the proton and theelectron to cross? b) How fast will each particle be going when their pathscross? c) How much time will elapse before the proton reaches theopposite plate? A pair of oppositely charged parallel plates are separated bya distance of 0.05m with a potential difference of 550V between theplates. A proton is released from rest at the positive plate at thesame time that an electron is released from rest at the negativeplate. Disregard any interation between the proton and theelectron. a) How long does it take for the paths of the proton and theelectron to cross? b) How fast will each particle be going when their pathscross? c) How much time will elapse before the proton reaches theopposite plate?

Explanation / Answer

We have the parallel plates as, V=Ed 550=E*0.05 E=11000N/C The force on each charge as F=Eq F=11000*1.6*10-19=1.76*10-15N The acceleration produced as F=ma For electron as 1.76*10-15=9.1*10-31*a For proton as 1.76*10-15=1.6726216*10-27*a We get the time taken for each as s=ut+1/2*at2 0.05=1/2*a*t2 We get their velocities as v=u+at v=at We get the time elapsed by their difference by as. Hence we get by it. We get their velocities as v=u+at v=at We get the time elapsed by their difference by as. Hence we get by it.

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