Question 449: Electrostatics & Capacitance

Question

The equivalent capacitance between plates ' A ' and ' B ' ( A -area of each plate, d-separation between plates) ( $\varepsilon_0$ - permittivity of free space) is

Accepted Answer

$$ 	ext { The equivalent circuit is given below } $$
$C_2$ and $C_3$ are connected in series, and this combination is then connected in parallel with $\mathrm{C}_1$.
$$ \begin{aligned} & C_1=\frac{K_1\left(\frac{A}{2}ight) \varepsilon_0}{d}=\frac{2 A \varepsilon_0}{d}=2 C \ & C_2=\frac{K_2\left(\frac{A}{2}ight) \varepsilon_0}{\left(\frac{d}{2}ight)}=\frac{3 A \varepsilon_0}{d}=3 C \ & C_3=\frac{K_3\left(\frac{A}{2}ight) \varepsilon_0}{\left(\frac{d}{2}ight)}=\frac{6 A \varepsilon_0}{d}=6 C \ & C_s=\frac{C_2 C_3}{C_2+C_3}=\frac{3 C 	imes 6 C}{3 C+6 C}=2 C \ & C_{e q}=C_1+C_s=2 C+2 C=4 C=\frac{4 A \varepsilon_0}{d} \end{aligned} $$

Answer

$\frac{\mathrm{A} \varepsilon_0}{\mathrm{~d}}$

Answer

$\frac{2 \mathrm{~A} \varepsilon_0}{\mathrm{~d}}$

Answer

$\frac{4 \mathrm{~A} \varepsilon_0}{\mathrm{~d}}$

Answer

$\frac{8 \mathrm{~A} \varepsilon_0}{\mathrm{~d}}$

Electrostatics & Capacitance

The equivalent capacitance between plates ' A ' and ' B ' ( A -area of each plate, d-separation between plates) ( \varepsilon_0 - permittivity of free space) is

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