Time varying Voltage

At the moment t = 0 t=0 , an electron leaves one plate of a parallel plate capacitor with a negligible velocity. An accelerating voltage varying as V = b t V = bt , where b b is constant, is applied between the plates. The separation between the plates is l l . Let the time taken by the electron to reach the opposite plate be t 0 t_0 in microseconds.

What is

t 0 ? \large \lfloor t_0 \rfloor \ ?

Details and Assumptions

b = 10 V / s b = 10 V/s

l = 1 m m l = 1 mm

e = 1.6 × 1 0 19 charge on electron e = 1.6 \times 10^{-19} \rightarrow \text{charge on electron}

m = 9.1 × 1 0 31 mass of electron m = 9.1 \times 10^{-31} \rightarrow \text{mass of electron}

Try my set


The answer is 1.

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2 solutions

Rahul Singh
Apr 15, 2017

we know that V = Ed....here d=l therefore - E = V/d = bt/l... now force on electron = eE = ma where a is the acceleration..... hence a = ebt/ml .... writing a as (dv/dt) = ebt/ml and integrating we get v(velocity) = (eb(t^2))/ml again writing v as (dx/dt) = (eb(t^2))/ml and integrating..... we get X = (eb(t^3))/6ml putting X = l ..we get t = 1.505 sec

Should be a mechanics question.....very easy though :P

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