Question

A charged particle is moving in an electric field of \(3 \times 10^{-10} \, \text{V} \, \text{m}^{-1}\) with mobility \(2.5 \times 10^{-6} \, \text{m}^2/\text{V}/\text{s}\) . Its drift velocity is

• A. \(2.5 \times 10^4 \, \text{m/s}\)
• B. 7.5 x 10^-4 m/s
• C. 1.2 x 10^-4 m/s
• D. 8.33 x 10^-4 m/s

Answer 1

The Correct Option is B

A charged particle moving in an electric field experiences a force that causes it to drift. The drift velocity (\(v_d\)) is the average velocity attained by the charged particle due to the electric field. The relationship between drift velocity, electric field (\(E\)), and mobility (\(\mu\)) is given by:

\(v_d = \mu E\)

We are given the following:

• Electric field: \(E = 3 \times 10^{-10} \text{ V/m}\)
• Mobility: \(\mu = 2.5 \times 10^{-6} \text{ m}^2/\text{V s}\)

Substituting these values into the formula, we can calculate the drift velocity:

\(v_d = (2.5 \times 10^{-6} \text{ m}^2/\text{V s}) \times (3 \times 10^{-10} \text{ V/m})\)

\(v_d = 2.5 \times 3 \times 10^{-6 - 10} \text{ m/s}\)

\(v_d = 7.5 \times 10^{-16} \text{ m/s}\)

Therefore, the drift velocity of the charged particle is:

7.5 x 10^-16 m/s