3 Marks Question

Question 13 Marks

A charge of +2.0 × 10^-8C is placed on the positive plate and a charge of -1.0 × 10^-8C on the negative plate of a parallel-plate capacitor of capacitance $1.2\times10^{-3}\mu\text{F}.$ Calculate the potential difference developed between the plates.

Answer

$\text{q}_1=+2.0\times10^{-8}\text{c}$
$\text{q}_2=-1.0\times10^{-8}\text{c}$
$\text{C}=1.2\times10^{-3}\mu\text{F}=1.2\times10^{-9}\text{F}$
$\text{net q}=\frac{\text{q}_1-\text{q}_2}{2}=\frac{3.0\times10^{-8}}{2}$
$\text{V}=\frac{\text{q}}{\text{c}}=\frac{3\times10^{-8}}{2}\times\frac{1}{1.2\times10^{-9}}=12.5\text{V}$

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Question 23 Marks

Suppose, one wishes to construct a 1.0 farad capacitor using circular discs. If the separation between the discs be kept at 1.0mm, what would be the radius of the discs?

Answer

Let the radius of the disc = R

$\therefore$ Area $=\pi\text{R}^2$
$\text{C}=1\text{f}$
$\text{D}=1\text{mm}=10^{-3}\text{m}$
$\therefore\text{C}=\frac{\in_0\text{A}}{\text{d}}$
$\Rightarrow1=\frac{8.85\times10^{-12}\times\pi\text{r}^2}{10^{-3}}$
$\Rightarrow\text{r}^2=\frac{10^{-3}\times10^{12}}{8.85\times\pi}=\frac{10^9}{27.784}$
$=5998.5\text{m}=6\text{Km}$

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Question 33 Marks

A metal sphere of radius R is charged to a potential V:

Find the electrostatic energy stored in the electric field within a concentric sphere of radius 2R.
Show that the electrostatic field energy stored outside the sphere of radius 2R equals that stored within it.

Answer

$\text{Q}=\text{CV}=4\pi\in_0\text{R}\times\text{V}$
$\text{E}=\frac{1}2{}\frac{\text{q}^2}{\text{C}}$ $[\therefore$ 'C' in a spherical shell $=4\pi\in_0\text{R}]$
$\text{E}=\frac{1}{2}\frac{16\pi^2\in_0^2\times\text{R}^2\times\text{V}^2}{4\pi\in_0\times2\text{R}}=2\pi\in_0\text{RV}^2$ $[$'C' of bigger shell $=4\pi\in_0\text{R}]$

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Question 43 Marks

A large conducting plane has a surface charge density $1.0\times10^{-4}\text{Cm}^{-2}.$ Find the electrostatic energy stored in a cubical volume of edge 1.0cm in front of the plane.

Answer

$\sigma=1\times10^{-4}\text{c/m}^2$
$\text{a}=1\text{cm}=1\times10^{-2}\text{m}$
$\text{a}^3=10^{-6}\text{m}$
The energy stored in the plane $=\frac{1}{2}\frac{\sigma^2}{\in_0}=\frac{1}{2}\frac{(1\times10^{-4})^2}{8.85\times10^{-12}}$
$=\frac{10^4}{17.7}=564.97$
The necessary electro static energy stored in a cubical volume of edge 1cm infront of the plane $=\frac{1}{2}\frac{\sigma^2}{\in_0}\text{a}^3=265\times10^{-6}=5.65\times10^{-4}\text{J}$

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Question 53 Marks

When 1.0 × 10¹² electrons are transferred from one conductor to another, a potential difference of 10V appears between the conductors. Calculate the capacitance of the two-conductor system.

Answer

Given that
Number of electron = 1 × 10¹²
Net charge Q = 1 × 10¹² × 1.6 × 10^-19 = 1.6 × 10^-7C
$\therefore$ The net potential difference = 10L.
$\therefore$ Capacitance
$\text{C}=\frac{\text{q}}{\text{v}}$
$=\frac{1.6\times10^{-7}}{10}=1.6\times10^{-8}\text{F}.$

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Question 63 Marks

A parallel-plate capacitor having plate area 25cm² and separation 1.00mm is connected to a battery of 6.0V. Calculate the charge flown through the battery. How much work has been done by the battery during the process?

Answer

$\text{A}=25\text{cm}^2=2.5\times10^{-3}\text{cm}^2$

$\text{d}=1\text{mm}=0.01\text{m}$

$\text{V}=\text{6V},\ \text{Q}=1$

$\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{8.854\times10^{-12}\times2.5\times10^{-3}}{0.01}$

$\text{Q}=\text{CV}=\frac{8.854\times10^{-12}\times2.5\times10^{-3}}{0.01}\times6$

$=1.32810\times10^{-10}\text{C}$

$\text{W}=\text{Q}\times\text{V}=1.32810\times10^{-10}\text{C}\times6$

$=8\times10^{-10}\text{J}.$

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Question 73 Marks

Find the charges on the three capacitors connected to a battery as shown in figure. Take $\text{C}_1=2.0\mu\text{F},\ \text{C}_2=4.0\mu\text{F},\ \text{C}_3=6.0\mu\text{F}$ and V = 12 volts.

Answer

$\text{C}_1=2\mu\text{F},\ \text{C}_2=4\mu\text{F},\ \text{C}_3=6\mu\text{F}$

$\text{V}=12\text{V}$

$\text{cq}=\text{C}_1+\text{C}_2+\text{C}_3$

$=2+4+6=12\mu\text{F}$

$=12\times10^{-6}\text{F}$

$\text{q}_1=12\times2=24\mu\text{C}$

$\text{q}_2=12\times4=48\mu\text{C}$

$\text{q}_3=12\times6=72\mu\text{C}$

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Question 83 Marks

Two capacitors of capacitances $4.0\mu\text{F}$ and $6.0\mu\text{F}$ are connected in series with a battery of 20V. Find the energy supplied by the battery.

Answer

$\therefore\text{C}_1=4\mu\text{F},\ \text{C}_2=6\mu\text{F},\ \text{V}=20\text{V}$

Eq. capacitor $\text{C}_{\text{eq}}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}=\frac{4\times6}{4+6}=2.4$

$\therefore$ The Eq Capacitance $\text{C}_{\text{eq}}=2.5\mu\text{F}$

$\therefore$ The energy supplied by the battery to each plate

$\text{E}=\Big(\frac{1}{2}\Big)\text{CV}^2$

$=\Big(\frac{1}{2}\Big)\times2.4\times20^2=480\mu\text{J}$

$\therefore$ The energy supplies by the battery to capacitor $=2\times480=960\mu\text{J}$

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Question 93 Marks

Find the charge appearing on each of the three capacitors shown in figure.

Answer

$\text{C}_1=8\mu\text{F},\ \text{C}_2=4\mu\text{F},\ \text{C}_3=4\mu\text{F}$

$\text{C}_{\text{eq}}=\frac{(\text{C}_2+\text{C}_3)\times\text{C}_1}{\text{C}_1+\text{C}_2+\text{C}_3}$

$=\frac{8\times8}{16}=4\mu\text{F}$

Since B & C are parallel & are in series with A.

So, $\text{q}_1=8\times6=48\mu\text{C}$

$\text{q}_2=4\times6=24\mu\text{C}$

$\text{q}_3=4\times6=24\text{C}\mu$

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Question 103 Marks

Each of the plates shown in figure has surface area $\Big(\frac{96}{\in_0}\Big)\times10^{-12}\text{Fm}$ on one side and the separation between the consecutive plates is 4.0mm. The emf of the battery connected is 10 volts. Find the magnitude of the charge supplied by the battery to each of the plates connected to it.

Answer

Here three capacitors are formed

And each of $\text{A}=\frac{96}{\in_0}\times10^{-12}\text{f.m.}$

$\text{d}=4\text{mm}=4\times10^{-3}\text{m}$

$\therefore$ Capacitance of a capacitor

$\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{\in_0\frac{96\times10^{-12}}{\in_0}}{4\times10^{-3}}=24\times10^{-9}\text{F}$

$\therefore$ As three capacitor are arranged is series

So, $\text{C}_\text{eq}=\frac{\text{C}}{\text{q}}=\frac{24\times10^{-9}}{3}=8\times10^{-9}$

$\therefore$ The total charge to a capacitor = 8 × 10^-9 × 10 = 8 × 10^-8c

$\therefore$ The charge of a single Plate = 2 × 8 × 10^-8 = 16 × 10^-8 = 0.16 × 10^-6 = 0.16μc.

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Question 113 Marks

Find the charge supplied by the battery in the arrangement shown in figure.

Answer

$\text{V}=\text{10v}$

$\text{C}_{\text{eq}}=\text{C}_1+\text{C}_2$ $[\therefore$ They are parallel$]$

$=5+6=11\mu\text{F}$

$\text{q}=\text{CV}=11\times10=110\mu\text{C}$

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Question 123 Marks

A charge of $20\mu\text{C}$ is placed on the positive plate of an isolated parallel-plate capacitor of capacitance $10\mu\text{F}.$ Calculate the potential difference developed between the plates.

Answer

$\therefore$ Given that

Capacitance $=10\mu\text{F}$

Charge $=20\mu\text{C}$

$\therefore$ The effective charge $=\frac{20-0}{2}=10\mu\text{F}$

$\therefore\text{C}=\frac{\text{q}}{\text{V}}$

$\Rightarrow\text{V}=\frac{\text{q}}{\text{C}}=\frac{10}{10}=1\text{V}$

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