An electron moves through a distance of 0.10 m parallel to the field lines of a uniform electric field of strength 2.0 kN C⁻¹ - AQA - A-Level Physics - Question 17 - 2017 - Paper 2

To find the work done (W) on the electron, we use the formula:

$W = F imes d$

where:

• W is the work done,
• F is the force applied,
• d is the distance moved in the direction of the force.

First, we need to calculate the force (F) acting on the electron in the electric field. The force on a charge in an electric field is given by:

$F = q imes E$

where:

• q is the charge of the electron (approximately $-1.6 imes 10^{-19}$ C),
• E is the electric field strength (2.0 kN C⁻¹ = 2000 N C⁻¹).

Now substituting the values:

$F = (-1.6 imes 10^{-19} ext{ C}) imes (2000 ext{ N C}^{-1}) = -3.2 imes 10^{-16} ext{ N}$

Next, we can calculate the work done:

$W = (-3.2 imes 10^{-16} ext{ N}) imes (0.10 ext{ m}) = -3.2 imes 10^{-17} ext{ J}$

However, since the question only concerns the magnitude and the electron moves parallel to the field lines, we interpret the work done as zero in this scenario because the displacement is parallel to the field direction, leading to no net work done against the electric field.