Two parallel plates, each having area a = 3557 cm2 are connected to the terminals of a battery of voltage vb = 6 v as shown. the plates are separated by a distance d = 0.59 cm. the battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.18 cm). what is the energy stored in this new capacitor?

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

3.2x10^7