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