Question

A particle having specific charge $C$ starts from rest in a region where the electric field has a constant direction but varying with time as $E=E_0{t}^2$ in magnitude. Further, it is observed that the particle has acquired a velocity $V$ after covering a distance $x$.Then

• A. $x=\frac{E_{0}c{t}^4}{12}$
• B. $x=\frac{E_{0}c{t}^4}{3}$
• C. $v=\frac{E_{0}c{t}^3}{4}$
• D. $v=\frac{E_0}{3}c{t}^3$

Answer 1

Answer: (A), (D)

The correct options are
A $x=\frac{E_{0}c{t}^4}{12}$
D $v=\frac{E_0}{3}c{t}^3$

$E=E_0{t}^2$
$\text{Since},a=\frac{qE}{m}=\frac{q{E}_0{t}^2}{m}=E_{0}c{t}^2$
$\Rightarrow \frac{dv}{dt}=E_{0}c{t}^2\Rightarrow \underset{0}{\overset{v}{\int }}dv=\underset{0}{\overset{t}{\int }}{E}_{0}c{t}^2\Rightarrow v=\frac{E_{0}c{t}^3}{3}$
$\text{Also},v=\frac{dx}{dt}=\frac{E_0}{3}c{t}^3\Rightarrow dx=\frac{E_0}{3}c{t}^{3}dt$
$\Rightarrow \underset{0}{\overset{x}{\int }}dx=\underset{0}{\overset{t}{\int }}\frac{E_{0}c{t}^3}{3}\Rightarrow x=\frac{E_0}{3}\frac{c{t}^4}{4}$