A potential difference of 1600 V is established between two parallel plates 4 cm

Question

A potential difference of 1600 V is established between two parallel plates 4 cm apart. An electron is released from the negative plate at the same time that a proton is released from the positive plate. (a) How far will they be from the positive plate when they pass each other? (b) How do their velocities compare when they strike the opposite plate? (c) How do their energies compare when they strike the opposite plates?

Explanation / Answer

We know that E = V / y ae = qE / me ap = q E / mp (a) How far will they be from the positive plate when they pass each other? For electron Xe = y + 0.5 ae t^2 ........1 For proton Xp = 0.5 ap t^2 .......2 When both passes each other , then Xe = Xp ==> y + 0.5 ae t^2 = 0.5 ap t^2 ==> t^2 = 2y / (ap - ae) .......3 Solve eq 3 and substitute value of 't' in eq 2, you get the required answer. b) we know that KE = PE 0.5 mv^2 = qV ==> v = [2qV/m]^1/2 Therefore for electron ve = [2qV/me]^1/2 For proton vp = [2qV/mp]^1/2 Substitute values . c) KE = 0.5 mv^2 Therefore For electron KEe = 0.5 mve^2 For proton KEp = 0.5 mvp^2 Substitute values.

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