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Chapter 3

Q. 3.7

An infinitely large parallel plane plates are spaced 0.8 cm apart. The voltage at one of the plates is raised from 0 to 5 V in 1 ns at a uniform rate, with respect to the other. After duration, the potential difference between the plates is suddenly dropped to 0 V and remains the same thereafter. Find (i) the position of the electron, which started with zero initial velocity from the negative plate, when the potential difference drops to zero volt, (ii) the total time of transit of the electron from the cathode to the anode.


Verified Solution

The electric field intensity,

E=\frac{5t}{10^{-9}d }=-\frac{5t}{10^{-9} \times 1 \times 10^{-2} }=5\times 10^{11}t(for  0\lt t\lt t_{1} )

=0 (for  t_{1}\lt t\lt \infty )

The velocity of electron,    v_{x}=-\int \frac{qE_{x} }{m}dt+C (C  is  the  initial  velocity)

=\int 5\times 10^{11}\frac{q}{m}t dt

=5\times 10^{11}\frac{q}{m}.\frac{t^{2} }{2}

The distance travelled by the electron,

       d=\int v_{x} dt=\int 5\times 10^{11}\frac{q}{m}.\frac{t^{2} }{2}dt+C            (C=0)

=5\times 10^{11}\frac{q}{m}.\frac{t^{2} }{2}

(i) The position of the electron after 1 ns,

       d=(5\times 10^{11} )\cdot (1.76\times 10^{11} )\cdot \frac{(1\times 10^{-9} )^{3} }{6}

=14.7 \times 10^{-6}m=14.7\mu m

(ii) The rest of the distance to be covered by the electron = 0.8 cm – 14.7 mm = 0.799 cm Since, the potential difference drops to zero volt, after 1 ns, the electron will travel the distance of 0.799 cm with a constant velocity of

       v_{x}=5\times 10^{11}\frac{q}{m}.\frac{t^{2} }{2}=(5\times 10^{11} )\cdot (1.76\times 10^{11} )\cdot \frac{(1\times 10^{-9} )^{3} }{6}

=44\times 10^{3}m/s

Therefore, the time              t_{2}=\frac{d}{v_{x}} =\frac{0.799\times 10^{-2} }{44\times 10^{3} }1.816\times 10^{-7}s

The total time of transit of electron from cathode to anode=1\times 10^{-9} +1.816\times 10^{-7}

   =1.816\times 10^{-7}s