Explain the effect of dielectric on capacitance of parallel plate capacitor and obtain the formula of dielectric constant.
Explain the effect of dielectric on capacitance of parallel plate capacitor and obtain the formula of dielectric constant.
Let us have two large plates each of area A separated by a distance $d$. The charge on the plates is $\pm \mathrm{Q}$ corresponding to the charge density $\pm \sigma$.
When there is vacuum between the plates,
$\mathrm{E}_{0}=\frac{\sigma}{\epsilon_{0}}$
and the potential difference $V_{0}$.
$\therefore \mathrm{V}_{0}=\mathrm{E}_{0} d$
If $\mathrm{C}_{0}$ is the capacitance,
$\mathrm{C}_{0}=\frac{\mathrm{Q}}{\mathrm{V}_{0}}$ $=\frac{\sigma \mathrm{A}}{\mathrm{E}_{0} d} \quad[\because \mathrm{Q}=\sigma \mathrm{A}]$ $\mathrm{C}_{0}=\frac{\epsilon_{0} \mathrm{~A}}{d} \quad \ldots(1)$ $\left[\because \mathrm{E}_{0}=\frac{\sigma}{\epsilon_{0}} \Rightarrow \epsilon_{0}=\frac{\sigma}{\mathrm{E}_{0}}\right]$
Consider a dielectric inserted between the plates fully occupying the intervening region. So, the dielectric is polarised by the field and surface charge density $\pm \sigma_{\mathrm{P}}$ arise on the plates.
Hence, electric field between two plates,
$\mathrm{E}=\mathrm{E}_{0}-\mathrm{E}_{\mathrm{P}}$ $\therefore \mathrm{E}=\frac{\sigma}{\epsilon_{0}}-\frac{\sigma_{\mathrm{P}}}{\epsilon_{0}} \quad\left[\because \mathrm{E}_{\mathrm{P}}=\frac{\sigma_{\mathrm{P}}}{\epsilon_{0}}\right]$ $\therefore \mathrm{E}=\frac{\sigma-\sigma_{\mathrm{P}}}{\epsilon_{0}}$ and electric potential difference,
$\mathrm{V}=\mathrm{E} d$
$=\frac{\sigma-\sigma_{\mathrm{p}}}{\epsilon_{0}} \cdot d$
For linear dielectric $\sigma$ is proportional to $\mathrm{E}_{0}$. Hence $\sigma-\sigma_{\mathrm{P}}$ is also proportional to $\mathrm{E}$, so we can write $\sigma-\sigma_{\mathrm{P}}=\frac{\sigma}{\mathrm{K}}$
Where $K$ is a constant characteristic of dielectric.
$\therefore \mathrm{V}=\frac{\sigma d}{\epsilon_{0} \mathrm{~K}}=\frac{\mathrm{Q} d}{\mathrm{~A} \in_{0} \mathrm{~K}} \quad\left[\because \sigma=\frac{\mathrm{Q}}{\mathrm{A}}\right]$
The capacitance with dielectric between the plate is then,
$\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{A} \in_{0} \mathrm{~K}}{d}=\mathrm{K} \frac{\mathrm{A} \in_{0}}{d}$
$\therefore \mathrm{C}=\mathrm{KC}_{0}$
$\quad \ldots \text { (2) from equation (1) }$
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Match the pairs
Capacitor
Capacitance
$(A)$ Cylindrical capacitor
$(i)$ ${4\pi { \in _0}R}$
$(B)$ Spherical capacitor
$(ii)$ $\frac{{KA{ \in _0}}}{d}$
$(C)$ Parallel plate capacitor having dielectric between its plates
$(iii)$ $\frac{{2\pi{ \in _0}\ell }}{{ln\left( {{r_2}/{r_1}} \right)}}$
$(D)$ Isolated spherical conductor
$(iv)$ $\frac{{4\pi { \in _0}{r_1}{r_2}}}{{{r_2} - {r_1}}}$
Match the pairs
| Capacitor | Capacitance |
| $(A)$ Cylindrical capacitor | $(i)$ ${4\pi { \in _0}R}$ |
| $(B)$ Spherical capacitor | $(ii)$ $\frac{{KA{ \in _0}}}{d}$ |
| $(C)$ Parallel plate capacitor having dielectric between its plates | $(iii)$ $\frac{{2\pi{ \in _0}\ell }}{{ln\left( {{r_2}/{r_1}} \right)}}$ |
| $(D)$ Isolated spherical conductor | $(iv)$ $\frac{{4\pi { \in _0}{r_1}{r_2}}}{{{r_2} - {r_1}}}$ |
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