i\'ve already done number 1 and 2 so i need the answers for 3,4 and 5. please be

Question

i've already done number 1 and 2 so i need the answers for 3,4 and 5. please be careful with the units. thanks.

Two parallel plates, each having area A = 3540cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.52cm. You may assume (contrary to the drawing) that the separation between the plates is small compared to a linear dimension of the plate. What is C, the capacitance of this parallel plate capacitor? What is Q, the charge stored on the top plate of the this capacitor?. A dielectric having dielectric constant K = 4 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 3540 cm2 and thickness equal to half of the separation (= 0.26 cm). What is the charge on the top plate of this capacitor? What is U, the energy stored in this capacitor?

Explanation / Answer

3) 1/Ceq = 2/C(old) + 2/K*C(old)

= 5/2C(old)

Ceq = 2/5 *C(old)

= 0.4 * 6.0193 *10^-4 uF

voltage across top resistor , V1 = V*C2/C1+C2 = V*2C(old) / (2C(old) + C(old)/2)

=4/5 V

= 4/5 *6 = 4.8 V

current in top plate , Q=CV1 = C(old)/2 *4.8

= 6.0193*10^-4 /2 *4.8 uC

= 0.001444 uC

4) U = 1/2 Ceq V^2

= 0.5 * 0.4*6.0193*10^-4 *6^2 *10^-6

=43.33 *10^-10 J

5) C1V1 + C2V2 = CV

C(old)/2 *4.8 + 2*C(old)*1.2 = C(old)*V

V = 4.8 V

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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