Answer to Question #166043 in Electric Circuits for Krishna

Question

electrical engineering questions and answers / a parallel...

Question: A parallel plate capacitor has an area of 5.00 cm2 and a capacitance of 3.50 pF. The capacitor is...

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

Given quantities:

$S = 5cm^2 = 5*10^{-4}m^2$ $C = 3.5pF$ $U = 12V$ $\varepsilon_o = 8.85*10^{-12}$

a) $C = \large\frac{\varepsilon\varepsilon_oS}{d} \to$ $d = \large\frac{\varepsilon\varepsilon_oS}{C}$ $= \large\frac{1*8.85*10^{-12}*5*10^{-4}}{3.5*10^{-12}}$ = $12.64*10^{-4}m = 1.3mm$

$W = \large\frac{CU^2}{2} = \large\frac{3.5pF*144}{2} =$ $252pJ$

b) The voltage doesn’t change (it's maintained by the battery). $\varepsilon = 3.5$

$\sigma$  - charge density  $\sigma = \large\frac{q}{S}$$= \large\frac{C_1U}{S}$$= \large\frac{\varepsilon CU}{S}$ $= \large\frac{\varepsilon\varepsilon_oU}{d}$ $= \large\frac{8.85*10^{-12}*3.5*12}{12.64*10^{-4}}$ = $30 \large\frac{\mu C}{m^2}$