Energy changes upon insertion of a solid dielectric in a capacitor

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SUMMARY

The discussion focuses on the energy changes that occur when a solid dielectric is inserted into a parallel plate capacitor connected to a constant voltage source. It is established that during the charging process, half of the work done by the battery is stored as potential energy, while the other half is lost as heat or electromagnetic radiation. When a dielectric is partially immersed, the potential energy of the system changes due to charge transfer, and the attractive force on the dielectric is considered without accounting for energy losses. The key distinction is that while charging a capacitor at constant voltage, the average potential across the capacitor is only half of the source voltage, leading to energy dissipation in the battery or wires.

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  • Understanding of parallel plate capacitors
  • Knowledge of dielectric materials and their properties
  • Familiarity with energy conservation principles in electrical circuits
  • Basic concepts of voltage sources and their behavior
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High school students, physics enthusiasts, electrical engineering students, and anyone interested in understanding the energy dynamics of capacitors and dielectrics in electrical circuits.

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Consider a simple circuit consisting of a battery and a parallel plate capacitor .During the process of charging of capacitors, we come to learn that half of the work done by the battery is stored in the form of potential energy and half of it is lost as heat or electromagnetic radiation.

Consider another arrangement consisting of parallel plate capacitor with its terminals connected with a battery to maintain a constant potential difference V₀ between it's terminals. Now I partially immerse a solid dielectric into this capacitor. The dielectric will feel an attractive force. Due to its further penetration inside the dielectric the potential energy of the system changes due to transfer of charge to maintain a constant potential difference . To calculate the force on the dielectric by this method , why do we neglect any loss of heat and say that all the work done by battery leads to change in potential energy of the system + the change in kinetic energy of the dielectric slab (with no energy loss as heat or electromagnetic radiation), while we can't neglect the loss in the aforementioned case. Can you please explain this?

Thanks in advance.
 
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The difference is that when charging the capacitor with a constant voltage source, the potential across the capacitor varies, and is only half the voltage of the voltage source on average, so the difference must be dissipated in the battery or in the wires, while when adding charge to a capacior at constant voltage, the voltage source can have the same voltage as the capacitor.

BTW, You can beat the 50% efficiency when charging a capacitor if you use a buck/boost voltage regulator with a varying voltage.
 
Thanks a lot for your reply. Well I'm a high school student and I don't have much knowledge regarding the devices you've mentioned. I'm highly constricted within my course syllabi. I would be happy if you throw some light on energy losses on insertion of dielectric at constant voltage ( a DC voltage)
 

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