3 A dielectric having dielectric constant K 28 is now insert

3) A dielectric having dielectric constant K 2.8 is now inserted in between the plates of the capacitor as shown. The dielectric has area A 2610 cm2 and thickness equal to half of the separation0.235 cm) What is the charge on the top plate of this capacitor? d/2 008232 C Submit Your submissons .008232 Computed value: .008232 Submitted: Tuesday, February 16 at 9:13 PM Feedback: It looks like you\'ve calculated the charge that would be on the plate if the dielectric completely filled the space between the plates. How can you split up the capacitor when the dielectric fills only the bottom half? Draw a picture of the situation to see how the capacitor can be separated.

Solution

1.

equivalent capacitance

C=C_{without dielectric} +C_dielectric

C=e_oA/(d/2) +ke_oA/(d/2)

C=(k+1)e_oA/(d/2) =(2.8+1)*(8.85*10^-12)(2610*10^-4)/(0.235*10^-2)

C=3.735*10^-9 F

Taken Voltage V=6 Volts

Charge on the capacitor is

Q=CV =(3.735*10^-9)*6

Q=2.24*10^-8 C =0.0224 uC

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a)

C2 and C3 are in series

1/C₂₃ =1/C₂ + 1/C₃ = 1/2.7 + 1/6.3

C₂₃ =1.89 uF

C₂₃ and C₄ are in parallel ,so equivalent capacitance C_ab is

C_ab=C₂₃+C₄ =1.89+2.4

C_ab=4.29 uF

b)

C₁ and C_ab are in series ,so equivalent capacitance C_ac is

C_ac =1/C_ab + 1/C₁ =1/4.29 +1/3.2

C_ac =1.83 uF

c)

equivalent capacitance

1/C_eq =1/C₅ +1/C_ac =1/3.2 +1/1.83

C_eq =1.165 uF

Total Charge

Q=C_eqV =12*1.165

Q=13.98 uC =14 uC (approx)

Charge on C5 is

Q₅=Q=14 uC