Complete question:
A parallel-plate capacitor has plates with an area of 405 cm² and an air-filled gap between the plates that is 2.25 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery. (a) How much energy is stored in the capacitor? (b) The separation between the plates is now increased to 4.50 mm. How much energy is stored in the capacitor now?
Answer:
The energy stored in the capacitor when the plates is increased to 4.50 mm is 1.32 X 10⁻⁵ J
Explanation:
Given:
Area of the plates = 405 cm² = 405 X (10⁻²)² m²= 405 X 10⁻⁴m² = 0.0405m²
Energy stored in a capacitor = CV²/2
Where;
V is the voltage across the plates = 575 V
C is the capacitor =?
C = Kε(A/d)
K is constant = 1.0
ε is permittivity of free space = 8.885 X 10⁻¹²
d is the diameter of the two plates = 2.25 mm = 0.00225m
C = 1.0 x 8.885 X 10⁻¹² x (0.0405/0.00225)
C = 1.5993 X 10⁻¹⁰ F
(a) Energy stored in a capacitor = 0.5 X 1.5993 X 10⁻⁹ X 575²
= 2.64 X 10⁻⁵ J
(b) The separation between the plates is now increased to 4.50 mm.
C = Kε(A/d)
New diameter, d = 4.5 mm = 0.0045 m
C = 1.0 x 8.885 X 10⁻¹² x (0.0405/0.0045)
C = 7.9965 X 10 ⁻¹¹ F
Energy stored in a capacitor = 0.5 X 7.9965 X 10 ⁻¹¹ X 575²
= 1.32 X 10⁻⁵ J
Therefore, the energy stored in the capacitor when the plates is increased to 4.50 mm is 1.32 X 10⁻⁵ J
