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Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by the discharge of a 1.29-F capacitor charged to 56.9 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 425, with one possessing a dielectric constant of 979. Find the electric potential energy of the original capacitor when it is charged.

1 Answer

2 votes

Answer:

potential energy = 2.088 ×
10^(9) J

Step-by-step explanation:

given data

Capacitance C = 1.29 F

voltage V = 56.9 kV = 56.9 × 10³ V

dielectric constant k1 = 425

new dielectric constant k2 = 979

to find out

electric potential energy of the original capacitor when it is charged

solution

we will apply here potential energy of capacitor formula that is

potential energy = 0.5 × C× V² ......................1

here C is Capacitance and V is voltage so potential energy will be from equation 1

potential energy = 0.5 × C× V²

potential energy = 0.5 × 1.29 × (56.9 × 10³)²

potential energy = 2.088 ×
10^(9) J

User Andrew Dinmore
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