W3 A constant potential difference of 9.0 V is maintained between the terminals

Question

W3

A constant potential difference of 9.0 V is maintained between the terminals of a 0.27-F parallel-plate air capacitor.

Part A A sheet of Mylar is inserted between the plates of the capacitor, completely filling the space between the plates. When this is done, how much additional charge flows onto the positive plate of the capacitor? (Dielectric constant K of Mylar is 3.1 )

ANSWER 5.1 microC

IN NEED PART B&C

What is the total induced charge on either face of the Mylar sheet?

PART C

What effect does the Mylar sheet have on the electric field between the plates? Explain how you can reconcile this with the increase in charge on the plates, which acts to increase the electric field.

EXPLAIN

Explanation / Answer

here,

potential difference, Vd = 9 V
capacitace of parallel plate capacitor, c = 0.27 uF = 2.7*10^-7 F

capacitace in presence of dielectric , cd = (3.1*2.7*10^-7) F

Part A :
Charge when dielectric is absent, Q
Q = c*V
Q = 2.7*10^-7 * 9
Q = 2.43*10^-6 C

Charge when dielectric is Present, Qd
Qd = cd*V
Qd = 3.1*2.7*10^-7 * 9
Qd = 7.533*10^-6 C

Net increase
Qi = Qd - Q
Qi = (7.533*10^-6) - (2.43*10^-6)
Qi = 5.103*10^-6 C

as 1C = 10-^6uC Therefore,
Qi = 5.1 uC

Part B and C :
it will equal to Charge when dielectric is introduced
Qd = 5.103*10^-6 C
or
Q = 5.103 uC

Part D:
Since , E = Q/eo*A, ( Area od plate = A , eo = permittivity of space )

which shows Electric field is directly proportional to Charge on plates, so if charge inc then Electric Field increase.

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