Answer to Question #166126 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?

Answer 1

The capacitance of a parallel-plate capacitor is defined as

$$C=\epsilon_r \epsilon_0 \frac A d$$

where

• A is the area of the plates (5 cm²=0.0005 m²)
• d is the plate separation (m)
• $\epsilon_0$  is the vacuum permittivity (~8.85*10^-12 F/m)
• $\epsilon_r$ is the relative permittivity of air (1.0006)

Then the separation of the plates:

$$d=\epsilon_r \epsilon_0 \frac A C=1.0006(8.85\cdot10^{-12} \frac F m) \frac {0.0005 m^2} {3.50\cdot10^{-12} F}\approx 0.00127m=0.127cm$$

The amount of energy stored between the plates of a parallel-plate capacitor:

$$\frac 1 2 CU^2_{battery}=\frac 1 2 (3.50\cdot10^{-12} F) (12.0V)^2 \approx 2.5 \cdot 10^{-10} J$$