Question #165813

A parallel plate capacitor has an area of 5.00 cm2 and a capacitance of 3.50 pF. The capacitor is connected to a 12.0 V battery. After the capacitor is completely charged up, the battery is removed

a. What must be the separation of the plates and how much energy is stored in the electric field between the plates?

b. If a sheet of Mica with εr=3.5 is placed between the plates, how will the voltage across the plates and the charge density on the plates change?

c. If the sheet of Mica has an area of 3.00 cm 2 so that it does not cover the whole area of the plates, what would be the effective capacitance of this arrangement?

d. If the Mica sheet is removed and the plates are moved 1.00 mm farther from each other, what will be the voltage between the plates, and how much work will it take to move the plates apart?

Expert's answer

a. The separation:

Energy:

b. With the dielectric, the capacitance and charge will increase ε_{r} times.

The charge density will increase 3.5 times:

The voltage will be the same:

c. This arrangement will be equal to two capacitors (with air and mica) connected in parallel:

"C_a=\\frac{\\epsilon (A-A_m)}{d},\\space\\space C_m=\\frac{\\epsilon\\epsilon_r (A_m)}{d},\\\\\\space\\\\\nC=C_a+C_m=\\frac{\\epsilon}{d}[A+A_m(\\epsilon_r-1)]=8.78\\text{ pF}."

d. Since the capacitor is not connected to the voltage source, the *charge will remain the same*, while the new capacitance will be

and the new voltage will be

"V_n=\\frac{Q_0}{C_n}=V\\frac{(d+x)}{d}=27.12\\text{ V}."

The work:

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