Each shot of the laser gun favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by the discharge of a 1.81-F capactor charged to 73.9 kV. Rosa rightly reckons that she can enhance the effect of each laser pulse by increasing the electric potential energy of the charged capacitor. This she would do by replacing the capacitor's filling, whose dielectric constant is 435, with one possessing a dielectric constant of 943.Find the electric potential energy of the original capacitor when it is charged. (in Joules)Calculate the electric potential energy of the upgraded capacitor when it is charged. ( In Joules)

The electric potential energy stored in a capacitor, expressed in terms of the value of the capacitance C, and the voltage between its terminals V, is as follows:

[tex]U =\frac{1}{2}*C*V^{2}[/tex]

a) For the original capacitor, we can find directly U as follows:

[tex]U =\frac{1}{2}*1.81F*(73.9e3)^{2} V2 = 4.94e9 J[/tex]

U = 4.94*10⁹ J

b) Prior to find the electric potential energy of the upgraded capacitor, we need to find out the value of the capacitance C of this capacitor, which is identical to the original, except that has a different dielectric constant.

As the capacitance is proportional to the dielectric constant, we can write the following proportion:

ε₂ / ε₁ = [tex]\frac{943}{435}= \frac{C2}{C1} =\frac{Cx}{1.81F}[/tex]

[tex]Cx =\frac{1.81F*943}{435} = 3.92 F[/tex]

Once calculated the new value of the capacitance, as V remains the same, we can find the electric potential energy for the upgraded capacitor as follows:

[tex]U =\frac{1}{2}*3.92F*(73.9e3)^{2} V2 = 1.07e10 J[/tex]

⇒ U = 1.07*10¹⁰ J