Questions: A -50 nC charged particle is in a uniform electric field E=(10 V / m, east). An external force moves the particle 1.0 m north, then 5.0 m east, then 2.0 m south, and finally 3.0 m west. The particle begins and ends its motion with zero velocity. a. How much work is done on it by the external force? b. What is the potential difference between the particle's final and initial positions?

Solution Steps

We are given a charged particle in a uniform electric field and asked to find the work done by an external force and the potential difference between the particle's final and initial positions.

• Charge of the particle, \( q = -50 \text{ nC} = -50 \times 10^{-9} \text{ C} \)
• Electric field, \( \vec{E} = 10 \text{ V/m (east)} \)
• Displacement path: 1.0 m north, 5.0 m east, 2.0 m south, 3.0 m west

• Net displacement in the north-south direction: \( 1.0 \text{ m north} - 2.0 \text{ m south} = -1.0 \text{ m south} \)
• Net displacement in the east-west direction: \( 5.0 \text{ m east} - 3.0 \text{ m west} = 2.0 \text{ m east} \)

• Work done by the electric field, \( W = q \vec{E} \cdot \vec{d} \)
• Net displacement vector, \( \vec{d} = 2.0 \text{ m east} - 1.0 \text{ m south} \)
• Since the electric field is only in the east direction, only the east component of displacement matters.
• Work done, \( W = q E d_{\text{east}} = (-50 \times 10^{-9} \text{ C}) (10 \text{ V/m}) (2.0 \text{ m}) = -1.0 \times 10^{-6} \text{ J} \)

• Potential difference, \( \Delta V = -\vec{E} \cdot \vec{d} \)
• Only the east component of displacement matters for potential difference.
• \( \Delta V = -E d_{\text{east}} = -(10 \text{ V/m}) (2.0 \text{ m}) = -20 \text{ V} \)

Final Answer