The distance between the plates of a charged parallel plate capacitor is $5\ cm$ and electric field inside the plates is $200\ Vcm^{-1}$. An uncharged metal bar of width $2\ cm$ is fully immersed into the capacitor. The length of the metal bar is same as that of plate of capacitor. The voltage across capacitor after the immersion of the bar is......$V$
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When the uncharged metal bar of width of $2 \mathrm{cm}$ is immersed into the capacitor, the distance between the plates of capacitor reduces and becomes: $(5-2) c m=3 c m$
since, the electric field$:$ $E=\frac{V}{d},$ where $E=200 \mathrm{V} / \mathrm{cm}, d=3 \mathrm{cm}$
so potential$:$ $V=E \times d=200 \times 3=600 V$
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