Question
Derive the formula $($equation$)$ of capacitance for a paralled plate capacitor.
✓
Answer
$\rightarrow$ A capacitor made up of two large parallel conducting plates kept at a small distance is called parallel plate capacitor.
$\rightarrow$ 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.
$\rightarrow$ Surface charge density on both the plates is $\sigma\left(=\frac{ Q }{ A }\right)$ and $-\sigma$ respectively.
$\rightarrow$ 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 $I , E ^{\prime}=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0$
$\rightarrow$ Electric field in the region below plate $II, E ^{\prime \prime}=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0$
$\rightarrow$ Electric field in the region above plate $I,$
$E ^{\prime}=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0$
$\rightarrow$ Electric field in the region below plate $\Pi$,
$E ^{\prime \prime}=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0$
$\rightarrow$ Electric field in the region between two plates,
$\therefore E =\frac{\sigma}{2 \varepsilon_0}+\frac{\sigma}{2 \varepsilon_0}$
$\therefore E =\frac{\sigma}{\varepsilon_0}$
$\therefore E =\frac{ Q }{\varepsilon_0 A}$
$\left(\because \sigma=\frac{ Q }{ A }\right)$
$\rightarrow$ Direction of this electric field is from $+ve$ plate to $-ve$ plate.
$\rightarrow$ The electric field is limited to the region between two plates and is uniform in that entire region.
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