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Energy stored by a capacitor Simplified Revision Notes

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7.4.3 Energy stored by a capacitor

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The energy stored in a capacitor (E)( E ) is the electrical energy that accumulates as electric charge builds up on its plates. This stored energy can be calculated by finding the area under a graph of charge (QQ) against potential difference (VV).

Energy Stored in a Capacitor

  1. Graphical Interpretation:
  • For a capacitor, charge (QQ) is directly proportional to potential difference (VV). This relationship produces a straight line through the origin when plotted on a graph, as shown in the diagram.
  • Since the area under this QQ -VV graph forms a right-angled triangle, we can calculate the area (and hence the energy stored) as:
E=12QVE = \frac{1}{2} QV image
  1. Alternate Forms of the Energy Equation:
  • By using the formula for capacitance C=QVC = \frac{Q}{V}, we can rearrange and substitute into the energy formula to find two other versions that may be useful depending on the given quantities.
  • These variations are:
E=12QV=12CV2=Q22CE = \frac{1}{2} QV = \frac{1}{2} CV^2 = \frac{Q^2}{2C}

Where:

  • QQ is the charge stored on the plates (in coulombs, CC),
  • VV is the potential difference across the plates (in volts, VV),
  • CC is the capacitance of the capacitor (in farads, FF).
  • EE is the energy stored (in joules, JJ),

Explanation of Each Form

  • E=12QVE = \frac{1}{2} QV: This form is most direct if both the charge and potential difference are known.

  • E=12CV2E = \frac{1}{2} CV^2 : Useful when you know the capacitance and potential difference across the capacitor.

  • E=Q22CE = \frac{Q^2}{2C}: Convenient if you have the charge and capacitance but not the potential difference.

infoNote

Example Problem

Example: A 200200 µFµF capacitor is charged to a potential difference of 1212 V. Calculate the energy stored in the capacitor.

Solution:

  1. Given:
  • Capacitance C=200μF=200×106FC = 200 \, \mu\text{F} = 200 \times 10^{-6} \, \text{F}
  • Potential difference V=12VV = 12 \, \text{V}
  1. Use the formula:
E=12CV2E = \frac{1}{2} CV^2
  1. Substitute the values:
E=12×(200×106)×(12)2E = \frac{1}{2} \times (200 \times 10^{-6}) \times (12)^2 E=12×0.0002×144E = \frac{1}{2} \times 0.0002 \times 144 E=0.0144JE = 0.0144 \, \text{J}

Answer: The energy stored in the capacitor is 0.0144J0.0144 \, \text{J} (or 14.4mJ14.4 mJ).

infoNote

Key Points

  • The energy stored in a capacitor is proportional to both the charge and the potential difference.
  • The formula for energy can be expressed in three forms depending on what values are known: E=12QVE = \frac{1}{2} QV, E=12CV2E = \frac{1}{2} CV^2, and E = Q22C\frac{Q^2}{2C}.
  • When analysing circuits with capacitors, use the form of the energy formula that best suits the given data.
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