Answer:
a. 1.63J
b. 22.27J
Explanation:
From the expression used in relating the capacitance of a capacitor to the area and distance stated below
[tex]C=\frac{E_{0}A}{d}[/tex]
from the question, the plates was not changed only the distance was varied. Since distance affect the capacitance, we can conclude that
[tex]E_{0}A=constant=K\\C=K/d\\Cd=k\\varrying k\\C_{1}d_{1}=C_{2}d_{2}=...=C_{n}d_{n}\\[/tex]
we can also conclude that the new capacitance will have a value of
[tex]C_{2}=\frac{C_{1}d_{1}}{d_{2}}[/tex]
[tex]\frac{C_{1}}{C_{2}}=\frac{d_{2}}{d_{1}}...........equation 1[/tex]
a. if the capacitor was disconnected from the potential source before the separation, we can conclude that the charge remain the same
Hence from
[tex]E=\frac{q^{2}}{2c}\\EC=q^{2}/2=k\\varrying k\\E_{1}C_{1}=E_{2}C_{2}\\E_{2}=\frac{E_{1}C_{1}}{C_{2}}\\[/tex]
earlier
[tex]\frac{C_{1}}{C_{2}}=\frac{d_{2}}{d_{1}}...........equation 1\\\frac{x}{y} Hence \\E_{2}=E_{1}*\frac{1}{3.7}\\E_{2}=6.02*0.27\\ E_{2}=1.63J[/tex]
b. if the capacitor remain connected before the separation, the voltage remain constant
Hence
[tex]E=\frac{CV^{2}}{2} \\\frac{E}{C}=\frac{V^{2}}{2}\\ \frac{E_{1}}{C_{1}}=\frac{E_{2}}{C_{2}}\\E_{2}= \frac{E_{1}}{C_{1}}C_{2}E_{2}= \frac{E_{1}}{d_{1}}d_{2}\\E_{2}=6.02*\frac{3.7}{1} \\E_{2}=22.27J[/tex]
