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Chapter 22: Problem 24

The classical picture of the hydrogen atom has the electron orbiting 0.0529 nm from the proton. What's the electric potential associated with the proton's electric field at this distance?

Short Answer

Expert verified
The electric potential associated with the proton's electric field at the distance of 0.0529 nm from the proton is approximately 2.717 Volts.

Step by step solution

01

Understanding The Task

The task requires us to calculate the electric field potential at a given distance from a proton. For this we need to understand the main concept, the electric potential formula, as well as the units of measurement.
02

Conversion of Distance to Meters

The distance given in the exercise is in nanometers (nm). However, the units in the electric potential formula are meters (m). So, first, we need to convert the given distance from nanometers to meters. One nanometer is equal to \( 1 \times 10^{-9} \) meters. Therefore, 0.0529 nm equals \( 0.0529 \times 10^{-9} \) meters, which simplifies to \( 5.29 \times 10^{-11} \) meters.
03

Calculation of The Electric Field Potential

Now that we have the distance in meters, we can plug all the numbers into the electric potential formula \( V = k_e \frac{Q}{r} \). Then we get \( V = (8.99 \times 10^9 \, N \cdot m^2/C^2) \cdot (1.602 \times 10^{-19} \, C/ 5.29 \times 10^{-11} \, m) \). Calculating this will give us the electric field potential in Volts.

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Most popular questions from this chapter

The electric potential in a region increases linearly with distance. What can you conclude about the electric field in this region?

The electric potential in a region is \(V=2 x y-3 z x+5 y^{2},\) with V in volts and the coordinates in meters. Find (a) the potential and (b) the components of the electric field at the point \(x=1 \mathrm{m}\) \(y=1 \mathrm{m}, z=1 \mathrm{m}\)

An electron passes point \(A\) moving at \(6.5 \mathrm{Mm} / \mathrm{s}\). At point \(B\) it's come to a stop. Find the potential difference \(\Delta V_{A B}\)

The electric field at the center of a uniformly charged ring is obviously zero, yet Example 22.6 shows that the potential at the center isn't zero. How is this possible?

How much work does it take to move a 50 - \(\mu\) C charge against a \(12-V\) potential difference?
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