Physics, 08.04.2020 02:35, michaellab1

An electron is placed at rest in the middle of this electric field (due to a parallel-plate capacitor). There is a vacuum between the plates. What can we say will happen to the electron while it is in this electric field

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answered: nuna37

This allows for a higher voltage rating than an air dielectric using a smaller total volume.

Explanation:

capacitors store the energy of the electrons in the form of an electrical charge on the plates the larger the plates and/or smaller their separation the greater will be the charge that the capacitor holds for any given voltage across its plates.

Electrons do not flow through a capacitor. They accumulate (as conduction band electrons) on the negative plate and deplete on the positive plate. This process continues as long as the voltage is applied and tapers off as the electric charge between the plates approaches the applied voltage.

In a capacitor, you have an electric field between the plates. This passes AC, but blocks DC, all without allowing electrons to pass between the plates. ... Indeed, at a high enough temperature (ballpark of 1000K) thermal energy alone is enough to 'cast' electrons out into the vacuum

answered: Guest

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

answered: Guest

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, $v = \dfrac{e_p}{q}$ then the following gives us the potential difference, $\delta v = v_b - v_a$ $\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}$ the formula $\delta v = \delta e_p / q$ may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get $\delta e_p = q\delta v$ using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get $\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}$ the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

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An electron is placed at rest in the middle of this electric field (due to a parallel-plate capacito...