Using the given constant electric-field magnitude 2 and current density J = x² over x=0 to 1, power is the integral of E·J: ∫₀¹2x²dx = 2/3.
Formula / Key Relation
pv = E·J
P = ∫₀¹ 2x² dx
= 2[x³/3]₀¹
= 2/3
Detailed Explanation
Concept & ReasoningElectromagnetic power density delivered to charges is p_v = E·J in W/m³. Interpreting the problem as a one-dimensional unit cross-section with E=2 and J=x², total power is obtained by integrating the power density along x from 0 to 1. The integral of x² is x³/3, so the result is 2/3 in the implied power units. The key is not to integrate J alone; the electric field factor must multiply it.How to Solve It in the ExamPower density in a conductor/field region = E·J.Important Exam PointCheck whether the problem implies unit area/volume when geometry is omitted.Related Revision PathThis question…
Concept & Reasoning
Electromagnetic power density delivered to charges is p_v = E·J in W/m³. Interpreting the problem as a one-dimensional unit cross-section with E=2 and J=x², total power is obtained by integrating the power density along x from 0 to 1. The integral of x² is x³/3, so the result is 2/3 in the implied power units. The key is not to integrate J alone; the electric field factor must multiply it.
How to Solve It in the Exam
Power density in a conductor/field region = E·J.
Important Exam Point
Check whether the problem implies unit area/volume when geometry is omitted.
Related Revision Path
This question sits in the revision path Electromagnetic Fields → Energy and Power → Field Power Density .
Quick Trick
Power density in a conductor/field region = E·J.
Common Mistake
Calling the given 2 'energy' instead of field quantity makes the wording awkward; use the dimensional relationship intended by the options.
Exam Tip
Check whether the problem implies unit area/volume when geometry is omitted.
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