Question

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assume that gravitational effects are negligible. At a time 6.36 s 6.36 s after it is released, the particle has a kinetic energy of 6.65 × 10 − 10 J. 6.65×10−10 J. At what time t t after the particle is released has it traveled through a potential difference of 0.351 V

Answer 1

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=$q=-2.74\times 10^{-6} C$

Kinetic energy of particle=$K_E=6.65\times 10^{-10} J$

Initial time=$t_1=6.36 s$

Final potential difference=$V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage=$V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.