Question

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 $Q/m=9.6\times 10^{7}C/kg,E=5\times 10^{5}V/m,y=84cm,g=10m/s^2$

• A. 4.03 × 10⁸ m
• B. 4.03 × 10¹⁰ m
• C. 4.03 × 10^-10 m
• D. 4.03 × 10^-8 m

Answer 1

Answer: (A)

Suppose particle falls down a distance y in time t so
$y=\frac{1}{2}g{t}^2;t=\sqrt{\frac{2y}{g}}$
Now for x-axis
$x=\frac{1}{2}{a}_x{t}^2=\frac{1}{2}.\frac{QE}{m}.\frac{2y}{g}$
$=\frac{1}{2}\times \frac{9.6\times 10^7\times 5\times 10^5\times 2\times 84\times 10^{-2}}{10}$
$=403.2\times 10^{10}m=4.03\times 10^{8}m$