Energy stored in a Capacitor

$U_C=\frac{Q^2}{2C}=\frac{1}{2}C{V}^2=\frac{1}{2}Q.V$

Energy supplied by Source

U_S = Q. V.

Energy lost in form of heat = $\frac{1}{2}Q.V$

Energy density of the electric field

$u=\frac{1}{2}{\in }_0{E}^2$ in free space

$u=\frac{1}{2}K{\in }_0{E}^2$ in presence of dielectric medium ; K - dielectric const.

Force on a Dielectric in a Capacitor

$F=\frac{Q^2}{2C^2}\left(\frac{dC}{dx}\right)$

when source is connected

V = $\frac{Q}{C}$

$\Rightarrow$ $F=\frac{1}{2}{V}^2\frac{dc}{dx}$

Illustration 1. A 4$\mu$f capacitor is charged to 150 V and another 6$\mu$f capacitor is charged to 200 V. Then they are connected across each other. Find the potential difference across them. Calculate the heat produced.

Solution : 4$\mu$f charged to 150 V would have q₁ = C₁V₁ = 600$\mu$C

6mf charged to 200 V would have q₂ = C₂V₂ = 1200$\mu$C

After connecting them across each other, they will have a common potential difference V.

Charges will readjust as q₁' and q₂'

$\begin{align} & V=\dfrac{{{q}_{1}}'}{{{C}_{1}}}=\dfrac{{{q}_{2}}'}{{{C}_{2}}}=\dfrac{{{q}_{1}}'+{{q}_{2}}'}{{{C}_{1}}+{{C}_{2}}}=\dfrac{{{q}_{1}}+{{q}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\dfrac{1800\mu C}{(4+6)\mu f}[conservation\,\,of\,\,ch\arg e] \\ & V=180\,\,volt. \\ & Initial\text{ energy} \\ & {{U}_{i}}=\dfrac{1}{2}{{C}_{1}}{{V}_{1}}^{2}+\dfrac{1}{2}{{C}_{2}}{{V}_{2}}^{2}\,\,\,\,\,\,\,=\dfrac{1}{2}(4\mu f){{(150V)}^{2}}+\dfrac{1}{2}(6\mu f).{{(200V)}^{2}}\,\,\,\,\,\,\,=\,\,0.165\,J \\ \end{align}$

Final energy

\begin{align}  {{U}_{f}}=\dfrac{1}{2}\left[ {{C}_{1}}+{{C}_{2}} \right].{{V}^{2}} \\  \,\,\,\,\,\,\,=\dfrac{1}{2}(4\mu f+6\mu f).{{(180)}^{2}} \\ \end{align}

= 0.161J

Heat produced = |U_f - U_i| = 0.003 J