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The electric potential varies in space according to the relation $V = 3x + 4y$. A particle of mass $0.1\,\, kg$ starts from rest from point $(2, 3·2)$ under the influence of this field. The charge on the particle is $+1\,\, μC$. Assume $V$ and $(x, y)$ are in $S.I.$ $units$ . The time taken to cross the $x-$ axis is.....$s$
$20$
$40$
$200$
$400$
Solution
$\mathrm{E}_{\mathrm{x}}=-\frac{\delta \mathrm{V}}{\delta \mathrm{x}}=-3 \mathrm{\,Vm}^{-1}$
$\mathrm{E}_{\mathrm{y}}=-\frac{\delta \mathrm{V}}{\delta \mathrm{y}}=-4 \mathrm{\,Vm}^{-1}$
$a_{x}=\frac{q E_{1}}{m}=\frac{1 \times 10^{-6} \times 3}{0.1}$
$=-3 \times 10^{-5} \mathrm{\,ms}^{-2}$
$a_{y}=\frac{q E_{y}}{m}=\frac{1 \times 10^{-6} \times 4}{0.1}$
$=-4 \times 10^{-5} \mathrm{\,ms}^{-2}$
Time taken to cross the $X$ – axis
Using $s=u t+\frac{1}{2} a t^{2}$
$3.2=\frac{1}{2} \times 4 \times 10^{-5} \times t^{2}$
$t=400 \mathrm{\,s}$
