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Serhud [2]
2 years ago
6

The plates of an isolated parallel plate capacitor are separated by a distance d and carry charge of magnitude q. The distance b

etween the plates is then reduced to d/2. How is the energy stored in the capacitor affected by this change
Physics
1 answer:
aleksklad [387]2 years ago
4 0

The energy stored in the capacitor decreases to one-half (½) of its initial value when the distance between the plates is then reduced to d/2.

<h3>The energy stored in a capacitor.</h3>

Mathematically, the energy stored in a capacitor can be calculated by using this formula:

U = ½CV² or U = q²/2C

<u>Where:</u>

  • q is the charge.
  • C is the capacitance.
  • V is the potential difference.

Thus, we can logically deduce that the energy stored in the capacitor decreases to one-half (½) of its initial value when the distance between the parallel plate capacitors is reduced to d/2.

Read more on charge here: brainly.com/question/4313738

#SPJ1

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