A parallel-plate air capacitor has rectangular plates each of len… - Vidyadip

Question

A parallel-plate air capacitor has rectangular plates each of length $20 cm$ and breadth $10 cm$. The separation between the plates is $1 mm$.
(a) Calculate the potential difference between the plates if $1 \mu C$ charge is given to the capacitor.
(b) With the same charge of $1 \mu C$, if the separation between the plates is doubled, what is the new potential difference?
(c) Calculate the electric field between the plates.

✓

Answer

Data $: Q =1 nC =10^{-9} C , I =20 cm , b =10 cm , d =1 mm =10^{-3} m , \varepsilon_0=8.85 \times 10^{-12}$ $F / m , k =1$ (air)
Area of the plates, $A = lb =20 \times 10=200 cm ^2=0.02 m ^2$
(a) The capacitance of the capacitor,
$
\begin{aligned}
C & =\frac{A k \varepsilon_0}{d}=\frac{0.02 \times 1 \times 8.85 \times 10^{-12}}{10^{-3}} \\
& =1.77 \times 10^{-10} F
\end{aligned}
$
The potential difference between the plates,
$
V=\frac{Q}{C}=\frac{10^{-9}}{1.77 \times 10^{-10}}= 5 . 6 5 V
$
(b) $E=\frac{V}{d} \quad \therefore E=\frac{V_1}{d_1}=\frac{V_2}{d_2}$
$\therefore$ The new potential difference,
$
V _2= V _1 \times \frac{d_2}{d_1}=5.65 \times 2=11.3 V \left(\because \frac{d_2}{d_1}=2 \text {, by the data }\right)
$
(c) The electric field between the plates,
$
E =\frac{V}{d}=\frac{5.65}{10^{-3}}=5650 N / C
$

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