Physics, 03.03.2020 16:30 Funkyatayo

A dielectric material is inserted between the charged plates of a parallel-plate capacitor. Do the following quantities increase, decrease, or remain the same as equilibrium is reestablished?

1. Charge on plates (plates remain connected to battery)
2. Electric potential energy (plates isolated from battery before inserting dielectric)
3.Capacitance (plates isolated from battery before inserting dielectric)
4. Voltage between plates (plates remain connected to battery)
5. Charge on plates (plates isolated from battery before inserting dielectric)
6. Capacitance (plates remain connected to battery)
7. Electric potential energy (plates remain connected to battery)
8. Voltage between plates (plates isolated from battery before inserting dielectric)

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Answer from: babowmanjacob666

1. Charge on plates (plates remain connected to battery) increases.

2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases.

3.Capacitance (plates isolated from battery before inserting dielectric) increases.

4. Voltage between plates (plates remain connected to battery) remains the same.

5. Charge on plates (plates isolated from battery before inserting dielectric) remains the same.

6. Capacitance (plates remain connected to battery) increases.

7. Electric potential energy (plates remain connected to battery) increases.

8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases.

Explanation:

When a dielectric material is inserted between the plates of a capacitor, the capacitance is increase by a factor K, the dielectric constant.

C = KC_0

By the capacitance formula, the other factors change accordingly.

1. Charge on plates (plates remain connected to battery) increases, because charge and capacitance are directly proportional.

2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases, because potential is inversely proportional to capacitance, and potential energy is given by the following formula

$U = \frac{1}{2}CV^2$

3.Capacitance (plates isolated from battery before inserting dielectric) increases.

4. Voltage between plates (plates remain connected to battery) stays the same, because the voltage is applied by the battery.

5. Charge on plates (plates isolated from battery before inserting dielectric) remains constant. If the plates isolated from the battery, then the total charge is conserved.

6. Capacitance (plates remain connected to battery) increases when a dielectric is inserted.

7. Electric potential energy (plates remain connected to battery) increases, because when plates remain connected to battery the voltage remains the same. But the capacitance increases. Therefore, electric potential energy increases.

8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases, because voltage is inversely proportional to capacitance.

Answer from: Quest

where is the photo and the answer choices?

Answer from: Quest

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