Question

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 1.89 F capacitor charged to 59.9 kV . Rosa rightly reckons that she can enhance the effect of each laser pulse by increasing the electric potential energy of the charged capacitor. She could do this by replacing the capacitor's filling, whose dielectric constant is 435 , with one possessing a dielectric constant of 923 . Find the electric potential energy of the original capacitor when it is charged.

Answer 1

The electric potential energy of the original 1.89 F capacitor charged to 59.9 kV is approximately 3.389×10¹¹ joules, calculated using the formula U = 1/2 CV².

Step-by-step explanation:

The electric potential energy of the original capacitor charged to 59.9 kV with a capacitance of 1.89 F can be calculated using the formula U = 1/2 CV², where U is the electric potential energy, C is the capacitance, and V is the voltage.

So, the electric potential energy is:

U = 1/2 × 1.89 F × (59.9 × 10³ V)²
U = 0.5 × 1.89 × (59,900)²
U = 0.945 × (59,900)²
U = 0.945 × (3.588×10¹¹) J
U = 3.38910¹¹ J

Therefore, the electric potential energy of the original capacitor when fully charged is approximately 3.389×10¹¹ joules.

Answer 2

Energy stored in original capacitor is
`32.89* 10^6J`

Step-by-step explanation:

We have given capacitance
`C=1.89F`

Capacitor is charged by
`V=59.9KV=59.9* 10^3volt`

We to find energy stored in the capacitor when it is charged

Energy stored in original capacitor is equal to

`E=(1)/(2)CV^2`

`E=(1)/(2)* 1.89* (5.9* 10^3)^2=32.89* 10^6J`

So energy stored in original capacitor is
`32.89* 10^6J`