Answer to Question #166128 in Electrical Engineering for Sai

Question

A parallel plate capacitor has an area of 5.00 cm 2 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?

Answer 1

a. Find separation and energy:

$$C_0=\frac{\epsilon_0 A}{d},\\\space\\ d=\frac{\epsilon_0A}{C_0}=1.26\text{ mm}.\\\space\\ E_i=\frac12 C_0V^2=25.2·10^{-9}\text{ nJ}.$$

b. The new charge and capacitance will be 3.5 times greater, so

$$\frac{\sigma_1}{\sigma_0}=\frac{Q_1/A}{Q_0/A}=\frac{\epsilon_rQ_0}{Q_0}=\epsilon_r=3.5$$

The voltage:

$$V=\frac{Q_1}{C_1}=\frac{\epsilon_rQ_0}{\epsilon_rC_0}=V.$$

c. The effective capacitance of this arrangement is the same as that of two capacitors in parallel, one with air and one with mica:

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

d. The capacitor is not connected to the voltage source anymore, so, the charge will not change. The new capacitance:

$$C_n=\frac{\epsilon_0 A}{d+x},$$

Voltage:

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

Work:

$$W=E_f-E_i=\frac12 C_nV_n^2-E_i=594.4\text{ pJ}.$$