PHYSICS HIGH SCHOOL

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?

Answers

Answer 1
Answer:

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* 10^(-8)\ J

C = Charge = 3.74* 10^(-6)\ C

Voltage is given by

V=(K)/(C)\n\Rightarrow V=(3.17* 10^(-8))/(3.74* 10^(-6))\n\Rightarrow V=0.00847593582888\ V

We have the relation

(V_i)/(V_f)=((t_i)/(t_f))^2\n\Rightarrow t_f=\sqrt{(t_i^2V_f)/(V_i)}\n\Rightarrow t_f=\sqrt{(8.36^2* 0.401)/(0.00847593582888)}\n\Rightarrow t_f=57.5022228905\ s

The time taken is 57.5022228905 s
Answer 2
Answer:

Final answer:

To find the time after which the particle has traveled through a potential difference of 0.401 V, use the formula KE = q · V and solve for t.

Explanation:

To find the time after which the particle has traveled through a potential difference of 0.401 V, we need to use the relationship between kinetic energy, electric potential difference, and charge. The formula is given by:

KE = q · V

where KE is the kinetic energy, q is the charge, and V is the potential difference. Rearranging the formula, we have:

t = KE / (q · V)

Substituting the given values, we get:

t = (3.17 · 10-8) / ((-3.74 · 10-6) · 0.401)

Solving for t, we find that the particle has traveled through a potential difference of 0.401 V at a time of approximately 2.07 seconds.

Learn more about Electric potential difference here:

brainly.com/question/33395355

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