A parallel plate capacitor is charged. Then battery is removed now If the plates are pulled apart
Medium
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${\rm{C}} = \frac{{{ \in _0}{\rm{A}}}}{{\rm{d}}}$ $\quad \quad {\rm{d}} \uparrow \Rightarrow {\rm{c}} \downarrow $
${\rm{V}} = \frac{\sigma }{{{ \in _0}}}{\rm{d}}\quad \quad {\rm{d}} \uparrow \to {\rm{V}} \uparrow $
$\mathrm{Q}=$ constant
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