Question

A parallel-plate vacuum capacitor has 7.64 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.30 mm,a) what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed?b) What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?

Answer 1

a) The stored energy is 4.34 J

b) The stored energy is 13.59 J

Step-by-step explanation:

A capacitor stores energy in its electric field according to the following formula:

`U=(Q^(2) )/(2C)=(1)/(2)CV^2`

Where U is the stored energy, Q the charge on a plate, V the voltage between plates and C its capacity.

When we change the distance between the plates,capacity changes, as:

`C_{\mbox{parallel-plate in a vacuum}} ={\varepsilon_(0)  S \over d}`

`\varepsilon_(0)` and S (area of the plate) remain constant, but d (distance between changes) makes capacity vary.

a) Charge remains constant (as it is disconnected from the circuit) but capacity changes.

As only d changes, we express C as a function of d and we find the new internal energy:

`U=(Q^(2) )/(2C)= \frac{Q^(2) }{2{\varepsilon_(0)  S \over d}}=(Q^(2) )/(2\varepsilon_(0)  S)*d\\\mbox{As it remains constant:  }(Q^(2) )/(2\varepsilon_(0)  S)=A\\\\7.68J=A*2.3*10^(-3)m\\A=3339.13 Jm\\\\U=A*d= 3339.13 Jm*1.3*10^(-3)m=4.34J`

b) Voltage remains constant, but capacity changes:

Knowing that, expressing C as a function of d:

`U=(1)/(2) (\varepsilon_(0) S)/(d) V^2\\\\7.68J=(1)/(2) \varepsilon_(0)  S V^2*(1)/(2.3*10^(-3)m)\\\\ \mbox{As it remains constant:}\\(1)/(2)*\varepsilon_(0)  S V^2=B \\\\7.68J=(B)/(2.3*10^(-3)m)`

`7.68J*{2.3*10^(-3)m=B=0.017664 Jm\\`

Knowing the value of the constant, we calculate the new internal energy:

`U=(B)/(d)=(0.017664Jm)/(1.3*10^-3m)=13.59J`