Answer:
Potential Energy of capacitor is
![= 7.09* 10^9 J](https://img.qammunity.org/2020/formulas/physics/college/5t09u9bqa5twbb817pu0bmh0yut9t9jvfm.png)
Step-by-step explanation:
Given data:
Capacitance C = 1.69 F
Voltage
![V = 66.1 kV = 66.1 * 10^3 V](https://img.qammunity.org/2020/formulas/physics/college/dvuqhm3uq8v7vedv1siy8klqlqo10jkztc.png)
Initial dielectric constant
![k_1 = 489](https://img.qammunity.org/2020/formulas/physics/college/vm6m12z8yudi83v8xeh8rj6ggaqux9mkfr.png)
New dielectric constant
![K_2 = 951](https://img.qammunity.org/2020/formulas/physics/college/ei4u0nt1zaj6ztye8bosf055wsigfpbnj6.png)
a) potential energy of capacitor
![U = (1)/(2) cv^2](https://img.qammunity.org/2020/formulas/physics/college/cu6yljrbv2d7dykzoj0nemewto8apd3cvx.png)
![= (1)/(2) 1.69 (66.1* 10^3)^2](https://img.qammunity.org/2020/formulas/physics/college/9b7dgkbej69tu075f1v4j9yvfy2pp1vh3r.png)
![= 3.69* 10^9 J](https://img.qammunity.org/2020/formulas/physics/college/bmgmhvguw0nbncngyydcms1r52a1l83ykj.png)
b) Upgraded capacitance
![C' = c(k_2)/(k_1)](https://img.qammunity.org/2020/formulas/physics/college/xlmyhl5bq0ncouwtma36cw6hy9f8ogdjrz.png)
Potential Energy of capacitor is
![U' =(1)/(2) c' * v^2](https://img.qammunity.org/2020/formulas/physics/college/r5p0nguenqggxoag2m2w1z4e96b185o95y.png)
![= (1)/(2) 3.28 (66.1* 10^3)^2](https://img.qammunity.org/2020/formulas/physics/college/6c24joquo4yug0e1u8me2rs8fz2mglq1ca.png)
![= 7.09* 10^9 J](https://img.qammunity.org/2020/formulas/physics/college/5t09u9bqa5twbb817pu0bmh0yut9t9jvfm.png)