[SOLVED] Capacitor problem involving dielectrics?

Capacitor problem involving dielectrics?

Two 5.0×5.0 mm electrodes with a 0.10mm thick sheet of Mylar between them are attached to a 9.0V battery. Without disconnecting the battery, the Mylar is withdrawn. (Very small spacers keep the electrode separation unchanged.) Dielectric constant for Mylar is k=3.1.

1)What is the charge before and after the Mylar is withdrawn?

2)What is the potential difference before and after the Mylar is withdrawn?

3)What is the electric field before and after the Mylar is withdrawn?

Thanks so much for the help

Answer:

The Best Answer for this question

Parallel plate cap

C = ε₀εr(A/d) in Farads

ε₀ is 8.8542e-12 F/m

εr is dielectric constant (vacuum = 1)

A and d are area of plate in m² and separation in m

Q = CV

calculate the capacitance with above formula, with and without mylar.

Use Q = CV to calculate the charge (which doesn’t change)

Use V = Q/C

field is voltage/distance

edit

initial C = (8.85e-12)(3.1)((0.005²)/(0.0001)) = 6.86e-14 F

final C = above/3.1 = 2.21e-14 F

Q1 = CV = 6.86e-14 x 9 = 6.17e-13 C (1)

Q2 = CV = 2.21e-14 x 9 = 1.99e-13 C

V initial = 9 volts

V final = 9 volts (2)

E = 9v/0.0001m = 9e-4 V/m, before and after

The battery stays connected, so the voltage must be the same across the capacitor. So the answer to 2) is no change, and the answer to 3) is also no change, since that’s just voltage/distance, and neither has changed.

What you want to do is calculate the new capacitance based on the dielectric constant. The spacing has not changed, and assuming that the dielectric constant of air is 1, the new capacitance will be 1/3.1 of the old. Since Q = CV, and the voltage didn’t change, then the new charge on the plates must be about 1/3 of what it was before.

Actually, when you pull the mylar out, a current will flow, and if that were integrated over time would equal to the charge lost.

This is the correct answer for this question Two 5.0×5.0 mm electrodes with a 0.10mm thick sheet of Mylar between them are attached to a 9.0V battery. Without disconnecting the battery, the Mylar is withdrawn. (Very small spacers keep the electrode separation unchanged.) Dielectric constant for Mylar is k=3.1.

2.) The potential difference is 9V before and 9V after

3.) E = V/d

= (9V) / (0.1*10^-3 m)

=90,000 V/m

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