Physics, 25.02.2020 03:35 dondre72

A parallel-plate capacitor has a plate separation of 1.5 mm and is charged to 600 V. If an electron leaves the negative plate, starting from rest, how fast is it going when it hits the positive plate?

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Answer from: LillianMRucker

GIven data:

Distance between the plates = 1.5 mm

Potential difference V = 600V

Charge on electron q = -1.6× $10^{-19}$ C

mass on electron = m = 9.1× $10^{-31}$ Kg

Solution:

First we will find the change in potential energy of the charge while moving through the potential difference of 600V.

ΔU = qΔV

= (-1.6× $10^{-19}$ )(600)

= -9.6× $10^{-19}$ J

By the law of conservation of mechanical energy, as there is no external force acting, so the sum of the kinetic and potential energies will be a constant.

K + U = E

ΔK + ΔU = 0

ΔK = -ΔU

1/2mv² = -ΔU

v² = -2ΔU/m

= $\frac{-2(-9.6*10^{-19}) }{9.1*10^{-31} }$

v = $\sqrt{2.11*10^{14} }$

v = 1.45× $10^{7}$ m/s

Answer from: brookekolmetz

1.45 x 10⁷m/s

Explanation:

In any system, the conservation of mechanical energy is always conserved. In other words, the sum of potential ( $E_{P}$ ) and kinetic energy ( $E_{K}$ ) is constant. i.e

$E_{P}$ + $E_{K}$ = k

=> Δ $E_{P}$ + Δ $E_{K}$ = 0

=> Δ $E_{P}$ = -Δ $E_{K}$ ---------------------(i)

The system considered here is an electric field.

(i) We know that the change in potential energy, Δ $E_{P}$ , of a charge Q as it moves between two points in an electric field is given as;

Δ $E_{P}$ = Q x Δ V ---------------(ii)

Where;

ΔV = change in electric potential as the charge moves negative to positive plate = 600V

Q = charge of an electron = -1.6 x 10⁻¹⁹C

Substitute these values into equation (ii)

Δ $E_{P}$ = -1.6 x 10⁻¹⁹C x 600V = -9.6 x 10⁻¹⁷J

(ii) Also, since the electron is starting from rest, the change in kinetic energy is given by;

$E_{K}$ = $\frac{1}{2}$ x m x v² -------------------(ii)

Where;

m = mass of the electron = 9.1 x 10⁻³¹ kg

v = final velocity of the electron

Substitute these values into equation (ii) as follows;

$E_{K}$ = $\frac{1}{2}$ x 9.1 x 10⁻³¹ x v² J

(iii) Now substitute the values of $E_{K}$ = $\frac{1}{2}$ x 9.1 x 10⁻³¹ x v² J and Δ $E_{P}$ = -9.6 x 10⁻¹⁷J into equation (i) as follows;

-9.6 x 10⁻¹⁷ = - $\frac{1}{2}$ x 9.1 x 10⁻³¹ x v²

2 x 9.6 x 10⁻¹⁷ = 9.1 x 10⁻³¹ x v²

v² = $\frac{2 * 9.6 * 10^{-17} }{9.1 * 10^{-31} }$

v² = 2.11 x 10¹⁴

v = $\sqrt{2.11 * 10^{14} }$

v = 1.45 x 10⁷ m/s

The electron is going as fast as 1.45 x 10⁷m/s

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