Name: Electric Flux Through a Closed Surface
Uploaded: 2019-07-19
Duration: 10 min 4 s
Description: In this video, let us understand the electric flux through a closed surface of a triangular box using Gauss's law of electrostatics.

Question

Find the equivalent capacitance between $P$ and $Q$ . Given, area of each plate $= A$ and separation between plates $= d$ .

Accepted Answer

(a) From the figure:

The capacitors are in series combination of three capacitors with plate areas $A$ and of plate separations $\frac{d}{3}$ $A$ .

Let $C_{1}, C_{2} & C_{3}$ be the 3 capacitors.

$C_{1}$ is filled with a dielectric of, $k_{1} = 1 .$

$C_{2}$ is filled with dielectric of $k_{2} = 3$ .

$C_{3}$ is filled with dielectric of $k_{3} = 6$ .

The capacitance of a parallel plate capacitor with a dielectric of dielectric constant $K$ is given by the formula,

$C = \frac{k ε_{0} A}{d}$

On substituting we get,

$C_{1} = \frac{k_{1} ε_{0} A}{\frac{d}{3}} = \frac{3 ε_{0} A}{d} k_{1} C_{2} = \frac{k_{2} ε_{0} A}{\frac{d}{3}} = \frac{3 ε_{0} A}{d} k_{2} C_{3} = \frac{k_{3} ε_{0} A}{\frac{d}{3}} = \frac{3 ε_{0} A}{d} k_{3}$

For capacitors connected in a series combination, the reciprocal of the equivalent capacitance is the sum of reciprocals of individual capacitances:

$\frac{1}{C_{s}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} ∴ \frac{1}{C_{s}} = \frac{1}{\frac{3 ε_{0} A}{d} k_{1}} + \frac{1}{\frac{3 ε_{0} A}{d} k_{2}} + \frac{1}{\frac{3 ε_{0} A}{d} k_{3}} \frac{1}{C_{s}} = \frac{d}{3 ε_{0} A k_{1}} + \frac{d}{3 ε_{0} A k_{2}} + \frac{d}{3 ε_{0} A k_{3}} \frac{1}{C_{s}} = \frac{d}{3 ε_{0} A k_{1}} + \frac{d}{3 ε_{0} A k_{2}} + \frac{d}{3 ε_{0} A k_{3}} \frac{1}{C_{s}} = \frac{d}{3 ε_{0} A} (\frac{1}{k_{1}} + \frac{1}{k_{2}} + \frac{1}{k_{3}}) \frac{1}{C_{s}} = \frac{d}{3 ε_{0} A} (\frac{k_{2} k_{3} + k_{1} k_{3} + k_{1} k_{2}}{k_{1} k_{2} k_{3}}) C_{s} = \frac{3 ε_{0} A}{d} (\frac{k_{1} k_{2} k_{3}}{k_{2} k_{3} + k_{1} k_{3} + k_{1} k_{2}})$

Substituting the values of we get,

$C_{s} = \frac{3 ε_{0} A}{d} (\frac{1 \times 3 \times 6}{3 \times 6 + 1 \times 6 + 1 \times 3}) C_{s} = \frac{3 ε_{0} A}{d} (\frac{18}{18 + 6 + 3}) C_{s} = \frac{3 ε_{0} A}{d} (\frac{18}{27}) C_{s} = \frac{ε_{0} A}{d} (\frac{18}{9}) C_{s} = \frac{2 ε_{0} A}{d}$

(b)From the figure:

The capacitors $C_{2}$ and $C_{3}$ , of plate areas $\frac{A}{2}$ and plate separations $\frac{d}{2}$ are connected in series.

This combination is in parallel with a capacitor $C_{1}$ of plate area $\frac{A}{2}$ and plate separation $d$ .

$C_{1}$ is filled with a dielectric of, $k_{1} = 4$ .

$C_{2}$ is filled with dielectric of $k_{2} = 3$ .

$C_{3}$ is filled with dielectric of $k_{3} = 6$ .

The capacitance of a parallel plate capacitor with a dielectric of dielectric constant $K$ is given by the formula,

$C = \frac{k ε_{0} A}{d}$

On substituting we get,

$C_{1} = \frac{k_{1} ε_{0} (\frac{A}{2})}{d} = \frac{ε_{0} A}{2 d} k_{1} C_{2} = \frac{k_{2} ε_{0} (\frac{A}{2})}{\frac{d}{2}} = \frac{ε_{0} A}{d} k_{2} C_{3} = \frac{k_{3} ε_{0} (\frac{A}{2})}{\frac{d}{2}} = \frac{ε_{0} A}{d} k_{3}$

For capacitors connected in a series combination, the reciprocal of the equivalent capacitance is the sum of reciprocals of individual capacitances:

$\frac{1}{C_{s}} = \frac{1}{C_{2}} + \frac{1}{C_{3}} ∴ \frac{1}{C_{s}} = \frac{1}{\frac{ε_{0} A}{d} k_{2}} + \frac{1}{\frac{ε_{0} A}{d} k_{3}} \frac{1}{C_{s}} = \frac{d}{ε_{0} A k_{2}} + \frac{d}{ε_{0} A k_{3}} \frac{1}{C_{s}} = \frac{d}{ε_{0} A k_{2}} + \frac{d}{ε_{0} A k_{3}} \frac{1}{C_{s}} = \frac{d}{ε_{0} A} (\frac{1}{k_{2}} + \frac{1}{k_{3}}) \frac{1}{C_{s}} = \frac{d}{ε_{0} A} (\frac{k_{3} + k_{2}}{k_{2} k_{3}}) C_{s} = \frac{ε_{0} A}{d} (\frac{k_{2} k_{3}}{k_{2} + k_{3}})$

Substituting the values of we get,

$C_{s} = \frac{ε_{0} A}{d} (\frac{3 \times 6}{6 + 3}) C_{s} = \frac{ε_{0} A}{d} (\frac{18}{9}) C_{s} = \frac{ε_{0} A}{d} (\frac{18}{9}) C_{s} = \frac{ε_{0} A}{d} (\frac{18}{9}) C_{s} = \frac{2 ε_{0} A}{d}$

$C_{s}$ and $C_{1}$ are in parallel

The equivalent capacitance of two capacitors connected in parallel is the sum of the individual capacitances.

$∴$ Finally, for the parallel combination of $C_{1}$ and $C_{s}$

$C' = C_{s} + C_{1} C' = \frac{2 ε_{0} A}{d} + \frac{ε_{0} A}{2 d} k_{1} C' = \frac{2 ε_{0} A}{d} + \frac{ε_{0} A}{2 d} (4) C' = \frac{4 ε_{0} A}{d}$

Thus, the equivalent capacitance $= \frac{4 ε_{0} A}{d}$ .

Find the equivalent capacitance between $P$ and $Q$ . Given, area of each plate $= A$ and separation between plates $= d$ .

Important Questions on Electrostatics

Learn and Prepare for any exam you want !

Find the equivalent capacitance between P and Q. Given, area of each plate = A and separation between plates = d.

Important Questions on Electrostatics

Learn and Prepare for any exam you want !

Find the equivalent capacitance between $P$ and $Q$ . Given, area of each plate $= A$ and separation between plates $= d$ .