Question
Derive the formula (/ equation) of capacitance for a paralled plate capacitor.
✓
Answer
→A capacitor made up of two large parallel conducting plates kept at a small distance is called parallel plate capacitor.
→Two parallel conducting plates are arranged parallel to each other as shown in figure. Area of each plate is A and perpendicular distance between the two plates is $d$. Charge on them is + Q and - Q respectively.
→Surface charge density on both the plates is $\sigma\left(=\frac{ Q }{ A }\right)$ and $-\sigma$ respectively.
→ Here the separation (d) between two plates is very small compared to the area of the plates. ( $d^2 \ll<$ A) Therefore, the electric field between the two plates can be considered uniform (So that we can use the formula $E=\frac{\sigma}{2 \varepsilon_0}$ to find out electric field due to both plates.) Electric field in the region above plate I , $E ^{\prime}=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0$
→Electric field in the region below plate II, $E ^{\prime \prime}=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0$
→Electric field in the region above plate I,
$E ^{\prime}=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0$
→Electric field in the region below plate $\Pi$,
$E ^{\prime \prime}=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0$
→Electric field in the region between two plates,
$\begin{array}{l}
\therefore E =\frac{\sigma}{2 \varepsilon_0}+\frac{\sigma}{2 \varepsilon_0} \\
\therefore E =\frac{\sigma}{\varepsilon_0} \\
\therefore E =\frac{ Q }{\varepsilon_0 A}
\end{array}$
$\left(\because \sigma=\frac{ Q }{ A }\right)$
→Direction of this electric field is from +ve plate to -ve plate.
→The electric field is limited to the region between two plates and is uniform in that entire region.
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