Question

A particle with charge − 3.74 × 10 − 6 C is released at rest in a region of constant, uniform electric field. Assume that gravitational effects are negligible. At a time 8.36 s after it is released, the particle has a kinetic energy of 3.17 × 10 − 8 J. At what time t after the particle is released has it traveled through a potential difference of 0.401 V?

Answer 1

Answer:

57.5022228905 s

Explanation:

$V_i$ = Initial voltage

$V_f$ = Final voltage = 0.401 V

$t_1$ = Initial time = 8.36 s

$t_2$ = Final time

K = Kinetic Energy = $3.18\times 10^{-8}\ J$

C = Charge = $3.74\times 10^{-6}\ C$

Voltage is given by

$V=\dfrac{K}{C}\\\Rightarrow V=\dfrac{3.17\times 10^{-8}}{3.74\times 10^{-6}}\\\Rightarrow V=0.00847593582888\ V$

We have the relation

$\dfrac{V_i}{V_f}=(\dfrac{t_i}{t_f})^2\\\Rightarrow t_f=\sqrt{\dfrac{t_i^2V_f}{V_i}}\\\Rightarrow t_f=\sqrt{\dfrac{8.36^2\times 0.401}{0.00847593582888}}\\\Rightarrow t_f=57.5022228905\ s$

The time taken is 57.5022228905 s