What potential difference is must be applied to produce an electric field that can accelerate an electron to 1/10 of velocity of light?

To determine the potential difference required to accelerate an electron to 1/10th the velocity of light, we can utilize the relationship between electric potential energy and kinetic energy. This involves some fundamental concepts from physics, particularly electromagnetism and relativistic mechanics.

Understanding the Basics

First, let's clarify a few key concepts:

• Electric Potential (Voltage): This is the work done per unit charge to move a charge from one point to another in an electric field.
• Kinetic Energy (KE): This is the energy an object possesses due to its motion, given by the formula KE = 1/2 mv², where m is mass and v is velocity.
• Relativistic Effects: At high speeds, particularly as we approach the speed of light, we need to consider relativistic effects, which modify the kinetic energy formula.

Calculating the Required Potential Difference

To find the potential difference (V) needed, we can set the electric potential energy equal to the kinetic energy gained by the electron. The electric potential energy (U) gained by an electron when it moves through a potential difference is given by:

U = eV

where e is the charge of the electron (approximately 1.6 x 10^-19 coulombs).

Next, for an electron accelerated to a speed of 1/10th the speed of light (c), we need to calculate its relativistic kinetic energy. The speed of light (c) is approximately 3 x 10⁸ m/s, so 1/10th of this speed is:

v = 0.1c = 0.1 x 3 x 10⁸ m/s = 3 x 10⁷ m/s.

In relativistic mechanics, the kinetic energy of an electron is given by:

KE = (γ - 1)mc²

where γ (gamma) is the Lorentz factor, defined as:

γ = 1 / √(1 - (v²/c²)).

Calculating γ for our speed:

v²/c² = (3 x 10⁷ m/s)² / (3 x 10⁸ m/s)² = 0.01,

thus:

γ = 1 / √(1 - 0.01) = 1 / √0.99 ≈ 1.005.

Now, substituting this into the kinetic energy formula:

KE = (1.005 - 1)(9.11 x 10^-31 kg)(3 x 10⁸ m/s)².

Calculating this gives:

KE ≈ 0.005 x (9.11 x 10^-31 kg)(9 x 10¹⁶ m²/s²) ≈ 4.59 x 10^-47 joules.

Relating Kinetic Energy to Potential Difference

Now we can set the electric potential energy equal to the kinetic energy:

eV = KE.

Substituting the values we have:

(1.6 x 10^-19 C)V = 4.59 x 10^-47 J.

Solving for V gives:

V = (4.59 x 10^-47 J) / (1.6 x 10^-19 C) ≈ 2.87 x 10^-28 volts.

Final Thoughts

This result indicates the potential difference required to accelerate an electron to 1/10th the speed of light is extraordinarily small, highlighting the efficiency of electric fields in accelerating charged particles. In practical terms, achieving such speeds in a controlled environment would involve sophisticated equipment and precise measurements, often found in particle accelerators.