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A parallel plate capacitor with a dielectric slab with a dielectric constant 3 filling the space betwen the plates is charged to a potential V. The battery is then disconnected and the dielectric slab is withdrawn. It is then replaced with another dielectric slab of dielectric constant 2. If the energies stored in the capacitor before and after the dielectric slab is changed are $E_1\: and E_2$, then $E_1 / E_2$ are

$\begin {array} {1 1} (1)\;\large\frac{9}{5} & \quad (2)\;\large\frac{4}{9} \\ (3)\;\large\frac{2}{3} & \quad (4)\;\large\frac{3}{2} \end {array}$

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(3) $\large\frac{2}{3}$
answered Nov 7, 2013 by pady_1
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