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A capacitor of capacitance \mathrm{C=900 pF} is charged fully by \mathrm{100V} battery B as shown in fig (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance \mathrm{C=900pF} as shown in fig(b). The electrostatic energy stored by system (b) is :

Option: 1

3.25 \times 10^{-6} \mathrm{~J} \\


Option: 2

2.25 \times 10^{-6} \mathrm{~J} \\


Option: 3

1.5 \times 10^{-6} \mathrm{~J} \\


Option: 4

4.5 \times 10^{-6} \mathrm{~J}


Answers (1)

( charged capacitor just connected                   (Steady state )

to uncharged capacitor )

As both the capacitor are in a parallel connection they attained the same potential at the steady state

Common potential   \mathrm{=V_{0} =\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}} \\

                             \mathrm{=\frac{100 C+0}{2 C}} \\

                             \mathrm{V_{0} =50 \mathrm{volt}}

\mathrm{U_{\text {initial }} =\frac{1}{2} C(V)^{2}=\frac{1}{2} \times 900 \times 10^{-12} \times 10^{4}} \\

              \mathrm{=450 \times 10^{-8} \mathrm{~J}}

\mathrm{U_{\text {final }} =\left(\frac{1}{2} C V_{0}^{2}+\frac{1}{2} C V_{0}^{2}\right) }\\

            \mathrm{=2\left(\frac{1}{2} \times 900 \times 10^{-9} \times(50)^{2}\right) }\\

            \mathrm{=225 \times 10^{-8} \mathrm{~J}}

Hence the correct option is 2

Posted by

Kshitij

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