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Q. A charge particle is moving along electric field as shown in figure. What is horizontal displacement of the charge particle when it descends a distance of $y$ meter?
Given $Q/m = 9.6 \times 10^7\, C/kg, E = 5 \times 10^5 V/m, y = 84 cm, g = 10 m/s^2$Physics Question Image

AIIMSAIIMS 2018Electric Charges and Fields

Solution:

Suppose particle falls down a distance y in time t so
$y=\frac{1}{2}gt^{2} ; t=\sqrt{\frac{2y}{g}}$
Now for x-axis
$x=\frac{1}{2}a_{x}t^{2}=\frac{1}{2}. \frac{QE}{m}. \frac{2y}{g}$
$=\frac{1}{2}\times\frac{9.6\times10^{7}\times5\times10^{5}\times2\times84\times10^{-2}}{10}$
$=403.2 × 10^{10} m = 4.03 × 10^{8} m$