A parrallel plate air capacitor of area A=11cm^2 and plate separation d=3.6mm is

Question

A parrallel plate air capacitor of area A=11cm^2 and plate separation d=3.6mm is charged by a battery to a voltage 68V 10. [1pt] A parallel-plate air capacitor of area A-11.0cm2 and plate separation d-3.60mm is charged by a battery to a voltage 68.0V If a dielectric material with kappa-4.60 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery), how much additional charge will flow from the battery onto the positive plate? Answer: 1.803E-13 C Incorrect, tries 3/6. Submit All Answers Last Answer: Hint: Calculate the change in capacitance, and then the change in charge. Watch out for the unit conversions.

Explanation / Answer

Remember and apply for parallel plate caapcitor, these as formulas

1. Capacitance C = K eoA/d  

where K is dieelctric constant   ( for air K = 1)

eo is permnittivity constant = 8.85 e-12

A is area = pi^2   ( for circular plates)    and   l* b   for rectanular plates

and

d is the distance between the plates



2.Charge Q = CV     where V is potrntial difference


3. energy stored U = 0.5 QV      or 0.5 CV^2    or 0.5 Q^2/C


4. electric field between the plates is E = Q/eo A    ( also given by E = V/d)

so here we now apply

initial charge Q = CV

where C is eoA/d

C = 8.85*10^-12 * 11*10^-4/(3.6 *10^-3)

C = 2.704pF

initial charge Q = 2.704pF * 68

Qi = 183.83 pC

now when K is inserted

C = KCo

C = 4.6 * 2.704

C = 12.43 pF

final charge Qf = 12.43 * 68 = 845.81 pC

so change of charge = Qf-Qi

= 845.81-183.83

Q = 661.98 pC is the change of charge -----<<<<<<<<ANSWER

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