A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V. (C is the capacitance and U is the stored energy.) Answer the following questions regarding the capacitor charged by a battery. For each statement below, select True or False.

1.) After being disconnected from the battery, inserting a dielectric with κ will decrease V.

2.) With the capacitor connected to the battery, inserting a dielectric with κ will increase C.

3.) After being disconnected from the battery, decreasing d decreases C.

4.) With the capacitor connected to the battery, inserting a dielectric with κ will decrease U.

5.) With the capacitor connected to the battery, decreasing d decreases Q.

6.) After being disconnected from the battery, decreasing d increases U.

Can't figure out why this is not correct.
1-T
2-F
3-T
4-F
5-F
6-F

Let's go through each statement and explain why each answer is either True or False.

1.) After being disconnected from the battery, inserting a dielectric with κ will decrease V.
Answer: False

Explanation: When a dielectric material is inserted between the plates of a capacitor, the capacitance (C) increases. The voltage (V) across the capacitor remains constant, as long as the charge (Q) is constant. Therefore, inserting a dielectric does not affect the voltage once it has been charged by the battery.

2.) With the capacitor connected to the battery, inserting a dielectric with κ will increase C.
Answer: True

Explanation: When a dielectric is inserted between the plates of a capacitor, it increases the capacitance (C). The capacitance is given by the equation C = κ * ε₀ * A / d, where κ is the dielectric constant, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between them. So, inserting a dielectric increases the capacitance.

3.) After being disconnected from the battery, decreasing d decreases C.
Answer: True

Explanation: The capacitance of a parallel plate capacitor is inversely proportional to the distance between the plates. So, as the distance d decreases, the capacitance C increases. When the capacitor is disconnected from the battery, the charge Q remains constant. Therefore, decreasing the distance between the plates will increase the capacitance.

4.) With the capacitor connected to the battery, inserting a dielectric with κ will decrease U.
Answer: False

Explanation: The energy stored in a capacitor (U) is given by the equation U = (1/2) * Q * V, where Q is the charge on the capacitor and V is the voltage across it. When a dielectric is inserted, the capacitance (C) increases, but the voltage (V) remains constant as long as the charge (Q) is constant. Therefore, the energy stored in the capacitor does not decrease.

5.) With the capacitor connected to the battery, decreasing d decreases Q.
Answer: False

Explanation: When a capacitor is connected to a battery, it charges up until the potential difference (V) across the plates matches the battery voltage. The charge (Q) on the capacitor is directly proportional to the capacitance (C) and the voltage (V). So, decreasing the distance between the plates (d) does not affect the charge Q while the capacitor is connected to the battery.

6.) After being disconnected from the battery, decreasing d increases U.
Answer: False

Explanation: The energy stored in a capacitor (U) is given by the equation U = (1/2) * Q * V, where Q is the charge on the capacitor and V is the voltage across it. When a capacitor is disconnected from the battery, the charge Q remains constant. The voltage V is also constant because it is determined by the charge on the capacitor and the capacitance. Therefore, decreasing the distance between the plates does not increase the energy stored in the capacitor.

I hope this clears up any confusion. Let me know if you have any more questions!