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)

Answers: 2

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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

Answer

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

explanation

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

Answer from: Quest

It's because 3 if it was 10 it wouldn't look right do it goes up to 4d

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A dielectric material is inserted between the charged plates of a parallel-plate capacitor. Do the f...

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