A charge of −3.0μC moves from a point at electric potential 250V to a point at 40V. Determine (a) the change in electric potential energy and (b) the work done by the electric field.
ΔV=40−250=−210V.
Then
ΔU=qΔV=(−3.0×10−6)(−210)=+6.30×10−4J.
Work done by the electric field is the negative of the potential-energy change:
Wfield=−ΔU=−6.30×10−4J.
Calculates potential difference.
[1 mark]
Calculates change in potential energy.
[2 marks]
Relates field work to potential-energy change.
[1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.