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 Fcapacitor charged to 60.9 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 431, with one possessing a dielectric constant of 947.

Required:
a. Find the electric potential energy of the original capacitor when it is charged. (in Joules)
b. Calculate the electric potential energy of the upgraded capacitor when it is charged. ( In Joules)

Question

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Respuesta :

okpalawalter8 okpalawalter8 · Answer 1 · 2020-09-20T02:17:42+02:00

Answer:

a

[tex]U = 3.505 *10^9 \ J[/tex]

b

[tex]U_1 = 7.696 *10^9 \ J[/tex]

Explanation:

From the question we are told that

The capacitance is [tex]C = 1.89 \ F[/tex]

The voltage is [tex]V = 60.9 \ k V = 60.9 *10^{3} \ V[/tex]

The first dielectric constant is [tex]\epsilon_1 = 431[/tex]

The second dielectric constant is [tex]\epsilon_2 = 947[/tex]

Generally the electric potential energy is mathematically represented as

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

=> [tex]U = \frac{1}{2} * 1.89 * (60.9 *10^{3})^2[/tex]

=> [tex]U = 3.505 *10^9 \ J[/tex]

Generally the capacitance when the capacitor's filling was changed is

[tex]C_n = 1.89 * \frac{947}{431}[/tex]

=> [tex]C_n = 4.15[/tex]

Generally the electric potential energy when the capacitor's filling was changed is

[tex]U_1 = \frac{1}{2} * C_1 * V^2[/tex]

=> [tex]U_1 = \frac{1}{2} * 4.15 * (60.9 *10^{3})^2[/tex]

=> [tex]U_1 = 7.696 *10^9 \ J[/tex]