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Question

a search m/ibiscms/mod/ibis/view.php?id=471 9187 ard UNT }, Mail-Kaur, Jaskira +UNT OrgSync MyUNTApple ing G Google Yahoo Jump to... Print Calculator Periodic Table stion 34 of 36 a Sapling Learning 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 1.25-F capacitor charged to 68.7 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 illing, whose dielectric constant is 475, with one possessing a dielectric constant of 931. Find the electric potential energy of the original capacitor when it is charged. Number Calculate the electric potential energy of the upgraded capacitor when it is charged. Number Previous e Check Answer Next tlE Hint 2

Explanation / Answer

We know that the electric potential energy of a capacitor is given by the equation

E = (CV2 )/2

where, C is the capacitance of the given capacitor and V is the potential given

in the given problem, C = 1.25 F and V= 68.7 kV =68700 V

hence the electric potential energy of the original capacitor is

E= (1.25)(68700)2 /2 = 2949806250 J = 2.9x109  J

the electric potential energy of the upgraded circuit is given by

E=(CV2)*(e1/e0)/2

changing the dielectric constant changes the capacitance by tht factor

here e1  =931 and e0 = 475

therefore

E =(1.25)(68700)2 (931/475) /2 = 5781620250 J = 5.78 x 109 J

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