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coop C Home | Chegg.comPre-lecture and hw f id Physics clickerNSCS OrgSync-NSCs Two parallel plates, each having area A 200m2- connocted to the terminals of a battery of voltage Vb as shown. The plates are separated by distance d You may assume (contrary to the drawing) that the 6Y dintiont of the plateo n tha plates is smal compared to a tinr 1) What is C, the capacitance of this parallel plate capacitor? 0.876 You currently have f submissions for this question. Only 3 submission are aflowed You can make 2 more submissions for this question Your submissions: 10.876 Computed value: 0.876 Submitted: Sunday, March at 4:38 PM Feedback 2) What is Q, the charge stored on the top plate of the this capacitor HC Submit You currently have O submissions for this question. Only 3 submission are allowed You can make 3 mare submissions for this question 3) A dielectric having dielectric constant 3.6 fs now inserted in between the plates of the capacitor as shown. The diolectric has area A 3203 cm2 and thickness equal to half of the separation (-0.25 cm) What is the charge on the top plate of this capacitor? d/2 You currentiy have O submissiors for this question Only 2 submission are ollowed. You can make 3 more submissions for this question 4) What is U, the energy stored in this capacitor? You currently have 0 submissions for this question Only 3 sutimissian are allowed You can make 3 more submissions for this questian 5) The battery is now disconnected from the capacitor and then the dielectric is withdrawn. What is V, the voltage across the capacitor?

Explanation / Answer

1)The capacitance of a capacitor is given by C = eo*A/d (where A is the area and d is the separation between the plates)

so,

C=8.85*10^-12*3201*10^-4/(0.005)

=5.66*10^-10 F

=5.66*10^-4 uF

2)Q=CV

=5.66*10^-4*6 uC

=33.96*10^-4 uC

3)C = eo*A/(d-t+t/K)

where t is the thickness of the dielectric and K is the dielectric constant.

so,

C = 8.85*10^-12*3201*10^-4/(0.005 - 0.0025 + 0.0025/3.6)

=8.86*10^-10 F

=8.86*10^-4 uF

so charge = CV

=8.86*10^-4*6

=53.16 *10^-4 uC

4)Energy = 0.5CV^2

=0.5*8.86*10^-4*6^2

=1.595*10^-8 J

5)The charge remains the same while the capacitance decreases. hence

Voltage = Q/C

=53.16/5.66

=9.392 V

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