Answer to Question #167081 in Electric Circuits for Vinit Kumawat

Question

A parallel plate capacitor has an area of 5.00 cm2 and a capacitance of 3.50 pF. The capacitor is connected to a 12.0 V battery. After the capacitor is completely charged up, the battery is removed

a. What must be the separation of the plates and how much energy is stored in the electric field between the plates?

b. If a sheet of Mica with εr=3.5 is placed between the plates, how will the voltage across the plates and the charge density on the plates change?

c. If the sheet of Mica has an area of 3.00 cm 2 so that it does not cover the whole area of the plates, what would be the effective capacitance of this arrangement?

d. If the Mica sheet is removed and the plates are moved 1.00 mm farther from each other, what will be the voltage between the plates, and how much work will it take to move the plates apart?

Answer 1

Area of plates, A = 5 cm² = 5 x 10^-4 m²

Capacitance, c = 3.50 pF = 3.50 x 10^-12 F

Voltage, V = 12V

(a) $c=\dfrac{\epsilon_{\omicron}A}{d}$

$\Rightarrow d=\dfrac{\epsilon_{\omicron}A}{c}$

$\Rightarrow d=\dfrac{8.85\times10^{-12}\times5\times10^{-4}}{3.50\times10^{-12}}$

$\Rightarrow d=12.64\times10^{-4}m$

Energy store, $E=\dfrac{1}{2}cV^2$

$\Rightarrow E=\dfrac{1}{2}\times3.50\times10^{-12}\times(12)^2$

$\Rightarrow E=242\times10^{-12}J$

(b) A dielectric medium, mica with relative permittivity $\epsilon_{r}=3.5$ is placed between the plates

$c=\dfrac{\epsilon A}{d}$

$\Rightarrow c= \dfrac{\epsilon_{\omicron}\epsilon_rA}{d}$

$\Rightarrow c= \dfrac{8.85\times10^{-12}\times3.5\times5\times10^{-4}}{12.64\times10^{-4}}$

$\Rightarrow c=12.25\times10^{-12} F$

Energy stored, $E=\dfrac{1}{2}cV^2$

$\Rightarrow V^2=\dfrac{2E}{c}$

$\Rightarrow V=\sqrt{\dfrac{2E}{c}}$

$\Rightarrow V=\sqrt{\dfrac{2\times252\times10^{-12}}{12.25\times10^{-12}}}$

$\Rightarrow V=6.414V$

Voltage of the battery decreases to 6.414V after placing mica sheet

$\dfrac{E_{\omicron}}{\epsilon_r}=E_{\omicron}-E_i$

$\Rightarrow \dfrac{\sigma_{\omicron}}{\epsilon_{\omicron}\epsilon_r}=\dfrac{\sigma_{\omicron}}{\epsilon_{\omicron}}-\dfrac{\sigma_i}{\epsilon_{\omicron}}$

$\Rightarrow \sigma_i=\sigma(1-\dfrac{1}{k})$

From above equation it is clear that $k=\epsilon_r$

When mica sheeet is placed between the plates, the charge density decreases.

(c) A₂ = 3 cm² = 3 x 10^-4 m²

$c^{'}=\dfrac{\epsilon_{\omicron}\epsilon_r A}{d}$

$\Rightarrow c^{'}=\dfrac{8.85\times10^{-12}\times3.5\times3\times10^{-4}}{12.64\times10^{-12}}$

$\Rightarrow c^{'}=7.35\times10^{-12}F$

(d) d' = 1mm = 1 x 10^-3 m

$c^{''}=\dfrac{\epsilon_{\omicron}A}{d}$

$\Rightarrow c^{''}=\dfrac{8.85\times10^{-12}\times5\times10^{-4}}{10^{-3}}$

$\Rightarrow c^{''}=4.425\times10^{-12}F$

$E=\dfrac{1}{2}c^{''}{V_1^{2}}$

$\Rightarrow V_1^{2}=\dfrac{252\times10^{-12}\times2}{4.425\times10^{-12}}$

$\Rightarrow V_1=10.670 V$

Work done = $\dfrac{1}{2}c^{''}V_1^{2}$

$=\dfrac{1}{2}\times4.425\times10^{-12}\times(10.670)^2$

$=251.93\times10^{-12}J$