Question

A parallel plate pacitor charged by a cell.After some time the cell is disconnectedand dielectric is inserted between the plate.How will affacted-1.Capacitance of capacitor.2.Potential difference 3.Energy stord in capacitor

Answer 1

ε0 –» Natural dielectric
ε–»Dilectric that is inserted
Q-charge
V-potential
C-capacitance.

$according \: to \: formula \\ c = \frac{q}{v} \\ and \\ c = \frac{Aε0}{d}$

A-Area between two plates of capacitors.

When dielectric of ε inserted then-

$c = \frac{Aε0ε}{d} \\ \\ means \: capacitance \: increase$
$ii)potential \: diffrence \\ v = \frac{q}{c} \\ c = \frac{Aε0}{d} \\ v = \frac{q}{ \frac{Aε0}{d} } = \frac{dq}{Aε0} \: \\ now \: on \: inserting \: dielectric \\ v = \frac{dq}{Aε0ε} \\ on \: compairing \: we \: get \: \: \: \\ that \: potential \: decreased$
$iii) \\ energy \: stored \: in \: capacitor = \frac{1}{2} c {v}^{2} \\ \\ since \: c \: is \: increased \: but \: v \: id \: decreased \: and \: it \: is \: squared \: so \\ \\ energy \: will \: also \: decrease$