Answer:
[tex] \Delta V=V_{2}-V_{1}=45.4V[/tex]
Explanation:
The energy, E, from a capacitor, with capacitance, C, and voltage V is:
[tex]E=\frac{1}{2} CV^{2}[/tex]
[tex]V=\sqrt{2E/C}[/tex]
If we increase the Voltage, the Energy increase also:
[tex]V_{1}=\sqrt{2E_{1}/C}[/tex]
[tex]V_{2}=\sqrt{2E_{2}/C}[/tex]
The voltage difference:
[tex]V_{2}-V_{1}=\sqrt{2E_{2}/C}-\sqrt{2E_{1}/C}[/tex]
[tex]V_{2}-V_{1}=\sqrt{2*55*10^{-6}/4.55*10^{-9}}-\sqrt{2*27.5*10^{-6}/4.55*10{-9}}=45.4V[/tex]
