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Firlakuza [10]
3 years ago
13

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

ative plate with a speed of 45000 m/s.
What will be the final speed of an electron released from rest at the negative plate?
Express your answer to two significant figures and include the appropriate units
V=_________
Physics
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:

2.1x10^6m/s

Explanation:

One electron has a charge of –1.602e-19 C

mass of electron is 9.1e-31 kg

mass of proton is 1.6726e−27 kg

mass ratio is 1.6726e−27 / 9.1e-31 = 1838

The force is constant, F

distance is constant, d

a = F/m

a increases by a factor 1838, as m decreases by that factor

a = a₀1838

v₀² = 2a₀d

v² = 2a₀d1838

v²/v₀² = 2a₀d1838 / 2a₀d = 1838

v² = 1838v₀² = 1838(45000)²

v = 45000√1838 = 2.1e6 m/s

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Explanation:

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Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec
Mnenie [13.5K]

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

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ΔU = -1.602 × 10⁻¹⁹ C × 142 V

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ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

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