Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilan

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 1.43-F capacitor charged to 69.1 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 475, with one possessing a dielectric constant of 971. ind the electric potn e thoriginal capacitor when it is charged Number Calculate the electric potential energy of the upgraded capacitor when it is charged. Number

Explanation / Answer

Consider a air filled capacitor of capacitance C , charge Q maintained at a potential difference V

Charge on the capacitor is Q = C * V

Consider a dielectric of dielectric constant K1 is introduced between the plates of the capacitor.

Then capacitance changes to C1 = K1 * C

Charge on the capacitor increases by a factor K1

Let it be Q1 = K1* Q

Introduction of a dielectric does not change the potential difference. So V1 = V

Then we can write Q1 = K1 * Q

Q1 = K1 * C * V

Q1 = C1 * V ......... ( 1 )

Similarly if some dielectric of dielectric constant K2 is introduced we can write

Q2 = K2 * C* V

Q2 = C2 * V .............. ( 2 )

From the equations ( 1 ) and ( 2 ) ,

Q1 / Q2 = C1 V / C2 V

Q1 / Q2 = C1 / C2

K1 Q / K2 Q = C1 / C2

C1 / C2 = K1 / K2

C2 = ( C1 * K2 ) / K1 ........ ( 3 )

Let this C1 = 1.43 F

V = 69.1 k V = 69.1 * 103 V

Energy stored in the capacitor C1 be U

U = 1/2 * K1 * C * V2

= 1/2 C1 * V2

= 1/2 * 1.43 * ( 69.1 × 103 )2

= 1/2 * 1.43 * 4774.81 * 106

= 3413.98 * 106 J

The energy stored in the original capacitor is U = 3413.98 M J

We have K1 = 475. , K2 = 971

When a dielectric is introduced , capacitance increases and so the energy stored in the capacitor.

When the dielectric of dielectric constant K2 is introduced , energy in the capacitor be U'

U' = 1/2 C2 V2

Using the value of C2 from the equation ( 3 ) , we can write

U' = 1/2 * ( C1 * K2 / K1 ) * V2

= 1/2 * ( 1.43 × 971 / 475 ) * ( 69.1 * 103 )2

= 6978.91 * 106 J

So the electrical potential energy stored in the upgraded capacitor is U' = 6978.91 * 106 J

Simple Method :

Energy stored in the capacitor C1 is

U = 1/2 C1 V2

= 1/2 * 1.43 * ( 69.1 * 103 )2

= 3413.98 * 106 J

Capacitance changes by a factor K2 / K1 and so the energy also changes by the same factor as potential difference does not change.

New energy is U' = ( K2 / K1 ) U

= ( 971 / 475 ) * 3413.98 * 106 J

= 2.044 * 3413.98 * 106 J

= 6978.91 * 106 J

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