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4 years ago

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An electric charge q of mass m in an oscillating electric field Eosinot experiences force q Eosinot. Suppose it starts from rest

at time to, and position x-0. a) What is its velocity as a function of time? b) What is its kinetic energy as a function of time? c) What is the particle's time averaged kinetic energy (averaged over an integer # of complete cycles)? d) The vector potential A is defined so that the electric field is E at and the canonical momentum is then given by Pcan P can (t). mv + qA. Find A(t) and

1 answer:

Masja [62]4 years ago

7 0

Answer:

speed of the charge electric is v = - (Eo q/m) cos t

Explanation:

The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,

F = q Eo sint

a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity

F = ma

q Eo sint = ma

a = Eo q / m sint

a = dv / dt

dv = adt

∫ dv = ∫ a dt

v-vo = I (Eoq / m) sin t dt

v- vo = Eo q / m (-cos t)

We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0

v = - (Eo q / m) cos t

b) Kinetic energy

K = ½ m v2

K = ½ m (Eoq / m)²2 (sint)²

K = ¹/₂ Eo² q² / m sin² t

c) The average kinetic energy over a period

K = ½ m v2

<v2> = (Eoq / m) 2 <cos2 t>

The average of cos2 t = ½, substitute and calculate

K = ½ m (Eoq / m)² ½

K = ¼ Eo² q² / m

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3 years ago

Read 2 more answers

Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the

anygoal [31]

Answer:

The value is the temperature of the air inside the tire $T_{2} =$ 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet $P_{1}$ = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet $P_{2}$ = 3.2 + 1 = 4.2 atm

Temperature at inlet $T_{1}$ = 300 K

(a) Volume of the system is constant so pressure is directly proportional to the temperature.

$\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }$

$\frac{T_{2} }{300} = \frac{4.2}{3.7}$

$T_{2} =$ 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

$m_{1} T_{1} = m_{2} T_{2}$

$\frac{m_{2} }{m_{1} } = \frac{T_{1} }{T_{2} }$

$\frac{m_{2} }{m_{1} } = \frac{300}{354.54}$

$\frac{m_{2} }{m_{1} } =0.00294$

value of the original mass of air in the tire should be released is $\frac{m_{2} - m_{1}}{m_{1}}$

$\frac{0.00294-1}{1}$

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

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