answers and explanations. Thank you A parallel plate capacitor with plate separa

Question

answers and explanations.

Thank you

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 energý.) Answer the following questions regarding the capacitor charged by à battery. For each statement below, select True or False False After being disconnected from the battery, increasing d increases U TrueWith the capacitor connected to the battery, increasing d decreasesQ TrueAfter being disconnected from the battery, inserting a dielectric with K will decrease U Faise After being disconnected from the battery, inserting a dielectric with k will increase V False After being disconnected from the battery, inserting a dielectric with will decrease C. True After being disconnected from the battery, decreasing d increases C.

Explanation / Answer

Energy in a Capacitor in Joules

E = ½CV² = ½QV = ½Q²/C

Q = CV

Parallel plate cap

C = (A/d) in Farads

is 8.854e-12 F/m

is dielectric constant (vacuum = 1)

A and d are area of plate in m² and separation in m

A) with the battery disconnected, Q is fixed. C decreases with increasing d. This decreases C. With both V and C changing, we need to use U = ½Q²/C. Decreasing C increases U. TRUE

B) V is fixed, C goes down. Q = CV. Q goes down. TRUE

C) C goes up, Q is fixed. U = ½Q²/C. U goes down. TRUE

D) FALSE

E) C goes up FALSE

F) C increases TRUE

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.