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Chapter 23: Problem 25

A proton, initially at rest, is accelerated through a potential difference of \(500 .\) V. What is its final velocity?

Short Answer

Expert verified
Answer: The final velocity of the proton is approximately \(3.09 \times 10^{5} m/s\).

Step by step solution

01

Identify the mass and charge of a proton

A proton has a mass of \(m_p = 1.67 \times 10^{-27} kg\) and a charge of \(q_p = 1.6 \times 10^{-19} C\).
02

Calculate the kinetic energy gained by the proton

Due to the potential difference, the proton will gain kinetic energy. The kinetic energy (KE) gained can be calculated using the formula: \(KE = q_p \times V\), where V is the potential difference. Plug in the values, we get: \(KE = (1.6 \times 10^{-19} C)(500V)\) \(KE = 8 \times 10^{-17} J\)
03

Use the kinetic energy formula to find the final velocity

The formula for kinetic energy is: \(KE = \frac{1}{2} m_p v^2\), where \(m_p\) is the mass of the proton and v is the final velocity. Let's solve for v using the obtained KE value: \(8 \times 10^{-17} J = \frac{1}{2} \times 1.67 \times 10^{-27} kg \times v^2\) Now, we can solve for v: \(v^2 = \frac{2 \times 8 \times 10^{-17}}{1.67 \times 10^{-27}}\) \(v^2 \approx 9.58 \times 10^{10}\) \(v \approx 3.09 \times 10^{5} m/s\)
04

Conclusion

The final velocity of the proton after being accelerated through a potential difference of 500 V is approximately \(3.09 \times 10^{5} m/s\).

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Most popular questions from this chapter

An electron moves away from a proton. Describe how the potential it encounters changes. Describe how its potential energy is changing.

An infinite plane of charge has a uniform charge distribution of \(+4.00 \mathrm{nC} / \mathrm{m}^{2}\) and is located in the \(y z\) -plane at \(x=0 . A+11.0 \mathrm{nC}\) fixed point charge is located at \(x=+2.00 \mathrm{~m}\) a) Find the electric potential \(V(x)\) on the \(x\) -axis from \(0

The electric potential energy of a continuous charge distribution can be found in a way similar to that used for systems of point charges in Section \(23.6,\) by breaking the distribution up into suitable pieces. Find the electric potential energy of an arbitrary spherically symmetrical charge distribution, \(\rho(r) .\) Do not assume that \(\rho(r)\) represents a point charge, that it is constant, that it is piecewise-constant, or that it does or does not end at any finite radius, \(r\). Your expression must cover all possibilities. Your expression may include an integral or integrals that cannot be evaluated without knowing the specific form of \(\rho(r) .\) (Hint: A spherical pearl is built up of thin layers of nacre added one by one.)

A ring with charge \(Q\) and radius \(R\) is in the \(y z\) -plane and centered on the origin. What is the electric potential a distance \(x\) above the center of the ring? Derive the electric field from this relationship.

A classroom Van de Graaff generator accumulates a charge of \(1.00 \cdot 10^{-6} \mathrm{C}\) on its spherical conductor, which has a radius of \(10.0 \mathrm{~cm}\) and stands on an insulating column. Neglecting the effects of the generator base or any other objects or fields, find the potential at the surface of the sphere. Assume that the potential is zero at infinity.
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