Question

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?

Answer 1

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