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So for a fixed charge, halving capacitance means double voltage, and voltage is a squared factor in energy.

So for a fixed charge, halving capacitance means double voltage, and voltage is a squared factor in energy.

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Russell McMahon
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$$ Q={C \over V} $$ $$ C={Q \over V} $$

$$ V={Q \over C} $$

$$ Q = C V $$

                              Energy:

$$ E={1 \over 2}C V^2 $$

$$ Q={C \over V} $$ $$ E={1 \over 2}C V^2 $$

$$ C={Q \over V} $$

$$ V={Q \over C} $$

$$ Q = C V $$

                              Energy:

$$ E={1 \over 2}C V^2 $$

Removed a confusing statement (can be rolled back if I'm wrong)
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Importantly - the generation methods do not contain reciprocal discharge mechanisms.
For discharge to occur the conditions for discharge need to be met separately. If they are not met - eg warm dry conditions with good insulation to other differentially charged objects then discharge will not occur. Charge will accumulate on the capacitor that the body forms relative to ground. For a given amount of charge, the lower the capacitance the higher the voltage. (Halve capacitance and voltage increases by 1/2th power.)

$$ Q={C \over V} $$ $$ E={1 \over 2}C V^2 $$

Importantly - the generation methods do not contain reciprocal discharge mechanisms.
For discharge to occur the conditions for discharge need to be met separately. If they are not met - eg warm dry conditions with good insulation to other differentially charged objects then discharge will not occur. Charge will accumulate on the capacitor that the body forms relative to ground. For a given amount of charge, the lower the capacitance the higher the voltage. (Halve capacitance and voltage increases by 1/2th power.)

Importantly - the generation methods do not contain reciprocal discharge mechanisms.
For discharge to occur the conditions for discharge need to be met separately. If they are not met - eg warm dry conditions with good insulation to other differentially charged objects then discharge will not occur. Charge will accumulate on the capacitor that the body forms relative to ground. For a given amount of charge, the lower the capacitance the higher the voltage.

$$ Q={C \over V} $$ $$ E={1 \over 2}C V^2 $$

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Russell McMahon
  • 152.1k
  • 18
  • 218
  • 402
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Source Link
Russell McMahon
  • 152.1k
  • 18
  • 218
  • 402
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