Answer:
a) The stored energy is 4.34 J
b) The stored energy is 13.59 J
Explanation:
A capacitor stores energy in its electric field according to the following formula:
[tex]U=\frac{Q^{2} }{2C}=\frac{1}{2}CV^2[/tex]
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:
[tex]C_{\mbox{parallel-plate in a vacuum}} ={\varepsilon_{0} S \over d}[/tex]
[tex]\varepsilon_{0}[/tex] 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:
[tex]U=\frac{Q^{2} }{2C}= \frac{Q^{2} }{2{\varepsilon_{0} S \over d}}=\frac{Q^{2} }{2\varepsilon_{0} S}*d\\\mbox{As it remains constant: }\frac{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[/tex]
b) Voltage remains constant, but capacity changes:
Knowing that, expressing C as a function of d:
[tex]U=\frac{1}{2} \frac{\varepsilon_{0} S}{d} V^2\\\\7.68J=\frac{1}{2} \varepsilon_{0} S V^2*\frac{1}{2.3*10^{-3}m}\\\\ \mbox{As it remains constant:}\\\frac{1}{2}*\varepsilon_{0} S V^2=B \\\\7.68J=\frac{B}{2.3*10^{-3}m}[/tex]
[tex]7.68J*{2.3*10^{-3}m=B=0.017664 Jm\\[/tex]
Knowing the value of the constant, we calculate the new internal energy:
[tex]U=\frac{B}{d}=\frac{0.017664Jm}{1.3*10^-3m}=13.59J[/tex]
