Question 472: Electrostats & Capacitance

Question

A parallel plate capacitor has plate area $50 \mathrm{~cm}^2$ and plate separation 3 mm . The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 4 . The capacitance becomes ( $\varepsilon_0=$ permittivity of free space)

Accepted Answer

Step 1: Recall concept
If a parallel-plate capacitor has partial dielectric filling in the direction of separation (i.e. layered dielectrics along thickness), then it is equivalent to capacitors
in series
.
Plate area: $A = 50 \,	ext{cm}^2 = 50 	imes 10^{-4}\, 	ext{m}^2 = 5 	imes 10^{-3}\, 	ext{m}^2$.
Separation: $d = 3 \,	ext{mm} = 3 	imes 10^{-3}\,	ext{m}$.
Thickness of dielectric: $t_d = 1 	imes 10^{-3}\,	ext{m}$. Relative permittivity: $\varepsilon_r = 4$.
Remaining thickness (air): $t_a = d - t_d = (3 - 1) \,	ext{mm} = 2 	imes 10^{-3}\,	ext{m}$.
So effectively we have:
A capacitor with dielectric slab: $\; t_d = 1 	imes 10^{-3}\,	ext{m}, \;\varepsilon = \varepsilon_0 \cdot 4$.
Another capacitor with air gap: $\; t_a = 2 	imes 10^{-3}\,	ext{m}, \;\varepsilon = \varepsilon_0$.
Both share same area $A$ and are in series.
Step 2: Capacitance of each section
Dielectric part:
$ C_1 = \frac{\varepsilon_0 \varepsilon_r A}{t_d}  = \frac{ 4 \varepsilon_0 \cdot 5	imes 10^{-3}}{1 	imes 10^{-3}}  = \frac{20 \varepsilon_0}{1}\,. $
So, $C_1 = 20 \varepsilon_0$.
Air part:
$ C_2 = \frac{\varepsilon_0 A}{t_a}  = \frac{ \varepsilon_0 \cdot 5	imes 10^{-3}}{2	imes 10^{-3}}  = \frac{5}{2}\varepsilon_0  = 2.5 \varepsilon_0\,. $
Step 3: Equivalent capacitance (series combination)
$ \frac{1}{C_	ext{eq}} = \frac{1}{C_1} + \frac{1}{C_2}. $
Substitute numbers:
$ \frac{1}{C_	ext{eq}} = \frac{1}{20 \varepsilon_0} + \frac{1}{2.5 \varepsilon_0}. $
$ = \frac{1}{20 \varepsilon_0} + \frac{1}{(5/2)\varepsilon_0} = \frac{1}{20 \varepsilon_0} + \frac{2}{5 \varepsilon_0}. $
$ = \frac{1}{20 \varepsilon_0} + \frac{8}{20 \varepsilon_0} = \frac{9}{20 \varepsilon_0}. $
Hence,
$ C_	ext{eq} = \frac{20}{9} \varepsilon_0. $
Final Answer:
$ \boxed{	frac{20 \varepsilon_0}{9}} $
That corresponds to
Option B
.

Answer

$\frac{18 \varepsilon_0}{7}$

Answer

$\frac{20 \varepsilon_0}{9}$

Answer

$\frac{16 \varepsilon_0}{7}$

Answer

$\frac{14 \varepsilon_0}{5}$

Electrostats & Capacitance

A parallel plate capacitor has plate area 50 \mathrm{~cm}^2 and plate separation 3 mm . The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 4 . The capacitance becomes ( \varepsilon_0= permittivity of free space)

Choose the correct answer: