Work done is the area under a voltage against charge graph

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SUMMARY

The discussion centers on calculating work done using the area under a voltage versus charge graph, specifically for a capacitor with a voltage of 4.5V and a capacitance of 9.0 microfarads. The key equations referenced include ΔW = VΔQ and the area of a triangle formula, 0.5 × b × h. The participant seeks clarification on how to articulate the relationship between energy stored in a capacitor and work done, emphasizing that the energy stored equals the work required to move charge onto the capacitor plates.

PREREQUISITES
  • Understanding of electric potential and charge relationships
  • Familiarity with capacitor energy storage concepts
  • Knowledge of basic calculus for area under curves
  • Ability to interpret voltage versus charge graphs
NEXT STEPS
  • Study the derivation of the energy stored in capacitors using ΔE = VΔQ
  • Learn how to calculate areas under curves in physics
  • Explore the implications of varying voltage in capacitor circuits
  • Review the principles of work and energy in electric fields
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism and capacitor behavior, as well as educators looking for clear explanations of energy storage in capacitors.

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Homework Statement



question_zpsa4cb5b1b.jpg


Homework Equations



Area of a triangle is 0.5 × b × h
ΔW=VΔQ

The Attempt at a Solution



I drew a straight line graph from the origin with end point 4.5V 9.0 microF
energy is a work done = .5QV


There is a mark in the mark scheme for ΔW=VΔQ explained.

What do you think I need to say for this mark?

any help gratefully received
 
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Your physics textbook, and perhaps your class lecture notes, should have an explanation of electric potential and how it relates to charge as well as work or energy. I would take a look at that.
 
Hi Redbelly thanks for responding.
I have just read in my notes that the energy stored in the capacitor is equal to the work done to force the extra charge onto the plates.

So ΔE=VΔQ
So V is changing and I guess if we make the increments very small in Q then the areas of each strip under the line add up to the total area under the triangle with base Q and Height V.
Thanks again
 

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