Question

$h8_plateA$

Two parallel plates, each having area A = 2082cm² are connected to the terminals of a battery of voltage V_b = 6 V as shown. The plates are separated by a distance d = 0.54cm. You may assume (contrary to the drawing) that the separation between the plates is small compared to a linear dimension of the plate.

1)What is C, the capacitance of this parallel plate capacitor?μF

2)What is Q, the charge stored on the top plate of the this capacitor?.μC

3)

A dielectric having dielectric constant κ = 2.7 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 2082 cm² and thickness equal to half of the separation (= 0.27 cm) . What is the charge on the top plate of this capacitor?μC

4)What is U, the energy stored in this capacitor?J

5)

$V=?$

The battery is now disconnected from the capacitor and then the dielectric is withdrawn. What is V, the voltage across the capacitor?

NEED HELP IN ALL PARTS PLEASE

Answer 1

solution : A = 2082 cm² VG = 60 d=0.54cm I co to A 8.8542x10"?x2082 410-4 a ł - 0.540102 C = 3.414 X10 F = 3.414x10 . MF to A. /E. A zk d) I kill ( 2. Q = CVb = 2.048X10-9C = 2048x10-3 uc 3. k= 27,t = d = 0.27 em c'= - Eod to A. d-t11-) c'= 4.982x10-10F = 4.982x104 uf d'= c'Vb = 2.989 4109 c = 2.989x103 uc 4. U = 1 / 2 = 1/2 d'Vo = 8.968x1095 5. charge remains constant. Q'= c'Vb = CV → V = (Ve) (25) Ik+ V = 8.756 V.