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02 Capacitor (2) Different Dielectrics

Aim

To show how the voltage of a charged capacitor changes when changing the dielectric.

Subjects

Diagram

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Figure 1:.

Equipment

Safety

*In accordance with IEC 61010-1 (Safety requirements for electrical equipment), a part is not deemed to be live (i.e. carrying a dangerous contact voltage) when, at voltages greater than extra-low voltage (> 60 V60 \mathrm{~V} DC), the current through an induction-free resistance of 2kOhm2 \mathrm{kOhm} is not greater than 2 mA2 \mathrm{~mA} for DC, additionally, the charge for voltages up to 15kV15 \mathrm{kV} is less than 45μC45 \mu \mathrm{C}, and the stored energy does not exceed 350 mJ350 \mathrm{~mJ} for voltages over 15kV15 \mathrm{kV}.

Presentation

The setup of the demonstration is explained to the students. The plates are set at such a separation that dd just a little bit larger than the thickness of the glass-plate. The power supply is set at 15kV15 \mathrm{kV}.

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Figure 2:.

The capacitor is charged by shortly touching the capacitor with the free lead of the 15kV15 \mathrm{kV} power supply (see Figure 2). After this charging of the capacitor, the voltmeter reads 15kV15 \mathrm{kV}.

The students are asked what will happen to the voltage of a charged capacitor when the glass plate is shifted between the plates (see Figure 3). After their answers shift the glass between the capacitor plates. They will see that the voltage lowers. Shift carefully all the time, with the glass plate sliding along the grounded plate of the capacitor, so the glass plate does not touch the high voltage positive plate!

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Figure 3:.

Removing the glass plate will increase the voltage again to its original value of 15kV15 \mathrm{kV}. The same experiment is performed with the container filled with water (see Figure 4).

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Figure 4:.

Explanation

Explanation 1:

The demonstration shows that the voltage lowers when a dielectric is shifted between the capacitor plates. Voltage is the energy per unit charge, so we can say that the potential energy of the capacitor lowers. Where is this energy gone? It was needed to polarize the dielectric! The “lost” energy is now in the polarized dipoles of the dielectric.

In the same way the energy stored in the capacitor becomes higher when the dielectric is removed. Also the demonstrator has to pull now because there is an attracting force between the induced charge on the dielectric and the plates. So he adds energy to the capacitor.

Explanation 2:

When the dielectric is between the plates, CC increases ( C=εAdC=\varepsilon \frac{A}{d} ) because ε\varepsilon becomes higher, and since QQ is constant this will cause VV to decrease (Q=CV)(Q=C V) .

(This explanation is just mathematics and no physics.)

Especially when water is the dielectric the voltage should lower dramatically, since εr\varepsilon_{r} of water is around 80 !

Remarks

Sources