Answer:
U_disconnected = 4.4383* 10^-8 J
Explanation:
- The complete information is missing, it is completed below.
Given:
- Two parallel plates with Area A = 3481 cm^2
- Connected to terminals of battery with Vb = 6 V
- The plates are separated by d = 0.52 cm
Find:
The battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.04 cm).
What is the energy stored in this new capacitor?
Solution:
- Calculate the total charge Q stored between the plates when the terminals were connected. The charge Q is given by:
Q = Vb*A*εo / d
Where, εo = 8.8542 * 10^-12 C / m-V ..... Permittivity of Free Space.
Q = 6*3481*8.8542 * 10^-12 / 0.52*100
Q = 7.3972*10^-9 C
- The total energy stored U in the capacitor is given by:
U = Q*Vb / 2
U = [ 7.3972*10^-9 ]* 6 / 2
U = 2.21915*10^-8 J
- Since the battery is first disconnected, it can do no work and so the charge on the plates remains constant. However moving the plate apart changes the voltage.
- Voltage is proportional to d, So if d doubles then Voltage doubles.
- Also Voltage and Total energy U is proportional so U also doubles.
- So , U:
U_disconnected = 2*U
U_disconnected = 2* 2.21915*10^-8
U_disconnected = 4.4383* 10^-8 J
