as you can see, i have already found the answer for 1 and 2, so i need the answe

Question

as you can see, i have already found the answer for 1 and 2, so i need the answers for 3,4,5 and 6. please be careful with the units(u can see what units is required for each section)

Two parallel plates, each having area A = 3481 cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.52 cm. What is Q, the charge on the top plate? What is U, the energy stored in the this capacitor? The battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.04 cm). What is the energy stored in this new capacitor? J What is E, the magnitude of the electric field in the region between the plates? Compare V, the magnitude of the new potential difference across the plates, to Vb, the voltage of the battery. V Vb Two uncharged parallel plates are now connected to the initial pair as shown. How will the electric field, E, and potential difference across the plates, V, Change, if at all? Both E and V will remain the same E will decrease and V will increase E will increase and V will decrease Both E and V will decrease Both E and V will increase

Explanation / Answer

c. when distance is changed to 1.04 cm

capaciatnce C = e0A/d

C = 8.85*10^-12 * 3481*10^-4/1.04*10^-2

C = 2.96*10^-10 F

eenrgy stored U = 0.5 CV^2

= 0.5 * 2.96*10^-10 * 6*6

= 5.3319*10^-9 J

d.E = Q/eoA

E = 3.55*10^-9/(8.85*10^-12* 3481*10^-4)

E = 1152.341 N/C

e. V>Vb

f. As EF E = Q/eoA

and V = Q/C = Qd/e0A

E will increases V will decrease

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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