Question

In the figure a capacitor of capacitance $2\mu F$ as is connected to a cell of eml 20 volt. The plates of the capacitors are drawn apart slowly to double the distance between them, The work done by the external agent on the plates is:

• A. $-200\mu J$
• B. $200\mu J$
• C. $400\mu J$
• D. $-400\mu J$

Answer 1

The correct option is D $-400\mu J$

When the plate separation is double, the charge remains constant and the capacitance is halved.The Work done in separating the plate is the difference in energy stored.

$\frac{Q^2}{2\left(\frac{C}{2}\right)}-\frac{Q^2}{2C}=-\frac{Q^2}{2C}$

$-\frac{{\left(CV\right)}^2}{2C}$

$-\frac{C{V}^2}{2}$

$=\frac{1}{2}\times 2\times {\left(20\right)}^2$

$=-400\mu J$