Answer:
U = 4.80 nJ
Explanation:
Given:-
- The area of two parallel plates, A = 3557 cm^2
- The battery has the p.d, Vb = 6 V
- The initial separation of plates, di = 0.59 cm
Find:-
he battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.18 cm).
What is the energy stored (U) in this new capacitor?
Solution:-
- The two parallel plates connected across a potential difference ( battery ) is considered as a capacitor with capacitance (C). The capacitance (C) can with (air in between) be calculated by the following formula:
C = ε*A / d
Where, ε : Permittivity of free space = 8.84 * 10^-12
- Solve for capacitance (C):
C = (8.84 * 10^-12)*( 0.3557 ) / (0.0059)
C = 5.32947 * 10^-10 F
- The energy stored by the capacitor (U) can be calculated from the following formula:
U = 0.5*C*V^2
U = 0.5*(5.32947 * 10^-10)*(6)^2
U = 9.59305 nJ
- When the plates are separated by double the initial distance we see that:
C ∝ 1 / d
When, d = 2*di
C = 0.5*C
- Also,
U ∝ C
Hence, C = 0.5 C
U = 0.5 U
So, The half of original energy stored calculated above is:
U = 0.5*9.59305 nJ
U = 4.80 nJ