Question

(1) A capacitor of 900 pF is charged with the help of 100 V battery. Calculate the electric potential energy of this capacitor. (2) The above capacitor is disconnected from the battery and is connected to another identical uncharged capacitor. What will be the total energy of the system ?

Answer 1

Answer:

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Answer 2

Answer:

(1)$4.5\times 10^{-6} J$     (2)$2.25\times 10^{-6}\ J$

Explanation:

(1) Given,

capacitance of the capacitor,

$c = 900\ pF = 900\times 10^{-12}\ F$

voltage of battery, v= 100 v

Energy of the capacitor,

$E\ =\ \dfrac{1}{2}.c.v^{2}$

$E\ =\ \dfrac{1}{2}\times 900\times 10^{-12}\times 100^{2}\ J$

          $=\ 4.5\times 10^{-6} J$

So, energy of the capacitor is$\ 4.5\times 10^{-6} J.$

(2) total charge on the capacitor,

  $q\ =\ c.v$

            $=\ 900\times 10^{-12}\times 100\ C$

                   $=\ 9\times 10^{-8}\ C$

Since, capacitor is disconnected from the battery and is connected to another identical uncharged capacitor so the equal amount of charge will be on both the capacitor.

i.e q1 = q2

   but total charge,

$q=q1+q2=\ 9\times 10^{-8}\ C$

$=> q1=q2=4.5\times 10^{-8}\ C$

So, total energy,

$E=\ \dfrac{1\times (q1)^2}{2.c1}+ \dfrac{1\times (q2)^2}{2.c2}$

        $=\ \dfrac{1\times (4.5\times 10^{-8})^2}{2\times900\times 10^{-12}}+ \dfrac{1\times (4.5\times 10^{-8})^2}{2\times900\times 10^{-12}}$

                $=2.25\times 10^{-6}\ J$

So, the total energy of the system will be $2.25\times 10^{-6}\ J$.