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

Explanations & Calculations

a)

• The capacitance of a capacitor with a vacuum in between the plates is given as $\small C= \large\frac{\epsilon_0A}{d}\small\cdots(1)$
• Therefore, using this the gap can be calculated

$\qquad\qquad \begin{aligned} \small 3.5\times 10^{-12}F&= \small \frac{8.85\times 10^{-12}Fm^{-1}\times5\times 10^{-4}m^2}{d}\\ \small d&= \small \bold{1.264\,mm} \end{aligned}$

• Energy stored is given by,

$\qquad\qquad \begin{aligned} \small E&= \small \frac{CV^2}{2}\\ &= \small \frac{3.5\times10^{-12}F\times(12V)^2}{2}\\ &= \small \bold{2.52\times10^{-10}J} \end{aligned}$

b)

• As a piece of a dielectric is inserted (assuming the entire gap is filled with Mica in this case), the capacitance increases as a result of the increment of the ability to hold more charges through the dielectric medium.
• Since the stored charge remains the same & the increased capacitance reduces the induced potential difference between the plates
• As the number of charges does not change, charge density remains constant.
• New capacitance is

$\qquad\qquad \begin{aligned} \small C_1&= \small \frac{\epsilon A}{d}\\ &= \small \frac{(\epsilon_r\cdot\epsilon_0)A}{d}\\ &= \small \epsilon_r\cdot(3.5\,pF)\\ &= \small \bold{12.25\,pF} \end{aligned}$

• New potential difference is,

$\qquad\qquad \begin{aligned} \small Q &=\small CV=C_1V_1\\ \small V_1&= \small \frac{3.5pF\times12V}{12.25pF}\\ &= \small \bold{3.429V} \end{aligned}$

c)

• Mica fills the entire gap (1.265mm) but partially the height between the plates ($\small 5cm^2/0.1264cm=39.56cm$)
• This arrangement results in a parallel combination of 2 capacitors: Mica filled one & air filled one
• Then the effective capacitance would be,

$\qquad\qquad \begin{aligned} \small C_{eff}&= \small C_{mica}+C_{air}\\ &= \small \frac{\epsilon_r\epsilon_0 A_1}{d}+\frac{\epsilon_0(A-A_1)}{d}\\ &= \small \frac{\epsilon_0}{d}\big[A+(\epsilon_r-1)A_1\big]\\ &= \small \frac{8.85\times10^{-14}Fcm^{-1}}{0.1264cm}\times\big[5cm^2+3(3.5-1)cm^2)\\ &= \small \bold{8.752\,pF} \end{aligned}$

d)

• According to te equation (1), capacitance is reduced by this act & according to $\small Q=CV$, the potential difference should increase accordingly.
• Now the separation is $\small 1.264\,mm+1\,mm=2.264\,mm$
• Then the new capacitance would be

$\qquad\qquad \begin{aligned} \small C_2& =\small \frac{\epsilon_0A}{d_1}\\ &= \small \frac{8.85\times10^{-14}Fcm^{-1}\times 5cm^2}{0.2264cm}\\ &= \small \bold{1.955\,pF} \end{aligned}$

• Then the new potential difference would be

$\qquad\qquad \begin{aligned} \small V_2 &= \small \frac{3.5pF\times12V}{1.955pF}\\ &= \small \bold{21.48V} \end{aligned}$

• Energy stored after the increament of the separation is

$\qquad\qquad \begin{aligned} \small E_1&= \small \frac{1.955\times 10^{-12}F\times (21.48V)^2}{2}\\ &= \small 4.51\times 10^{-10}J \end{aligned}$

• Then the needed external work to do thid modification is

$\qquad\qquad \begin{aligned} \small\Delta E&= \small E_1-E_0\\ &= \small (4.51-2.52)\times 10^{-10}J\\ &= \small \bold{2.26\times 10^{-10}J} \end{aligned}$