A capacitor with capacitance 5\; \mu F is charged to 5\; \mu C. If the plates are pulled apart to reduce the capacitance to 2\; \mu F, how much work is done ?

 

  • Option 1)

     6.25\times 10^{-6}J

  • Option 2)

     3.75\times 10^{-6}J

  • Option 3)

    2.16\times 10^{-6}J

  • Option 4)

     2.55\times 10^{-6}J

Answers (1)

 

W=\; uf-ui

       =\frac{Q^{2}}{2}\left [ \frac{1}{Cf}-\frac{1}{Ci} \right ]

      =\left ( \frac{5\mu C}{2} \right )^{2}\left [ \frac{1}{2\mu F}-\frac{1}{5\mu F} \right ]

     =\frac{25\times 10^{-6}}{2}\left [ \frac{3}{10} \right ]

      =\frac{7\; 5}{2}\times 10^{-7}

     =37.5\times 10^{-7}J

      =3.75\times 10^{-6}J


Option 1)

 6.25\times 10^{-6}J

Option 2)

 3.75\times 10^{-6}J

Option 3)

2.16\times 10^{-6}J

Option 4)

 2.55\times 10^{-6}J

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