Numerical Problems-1 - Chapter 1 - Electrostatics - 12th Physics Guide Samacheer Kalvi Solutions

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Numerical Problems
Question 1.
When two objects are rubbed with each other, approximately a charge of $50 \mathrm{nC}$ can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.
Solution:
Charge produced in each object $\mathrm{q}=50 \mathrm{nC}$
$
\mathrm{q}=50 \times 10^{-9} \mathrm{C}
$
Charge of electron $(\mathrm{e})=1.6 \times 10^{-9} \mathrm{C}$
Number of electron transferred, $\mathrm{n}=\frac{q}{e}=\frac{50 \times 10^{-9}}{1.6 \times 10^{-19}}$
$=31.25 \times 10^{-9} \times 10^{19}$
$\mathrm{n}=31.25 \times 10^{10}$ electrons
Ans. $\mathrm{n}=31.25 \times 10^{10}$ electrons
Question 2.
The total number of electrons in the human body is typically in the order of $10^{28}$. Suppose, due to some reason, you and your friend lost $1 \%$ of this number of electrons. Calculate the electrostatic force between you and your friend separated at a distance of $1 \mathrm{~m}$. Compare this with your weight. Assume mass of each person is $60 \mathrm{~kg}$ and use point charge approximation.
Solution:
Number of electrons in the human body $=10^{28}$
Number of electrons in me and my friend after lost of $1 \%=10^{28} \times 1 \%$
$=10^{28} \times \frac{1}{100}$
$\mathrm{n}=10^{26}$ electrons
Separation distance $\mathrm{d}=1 \mathrm{~m}$
Charge of each person $\mathrm{q}=10^{26} \times 1.6 \times 10^{-19}$
$q=1.6 \times 10^7 \mathrm{C}$

Electrostatic force, $\mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}=\frac{9 \times 10^9 \times 1.6 \times 10^7 \times 1.6 \times 10^7}{1^2}$
$\mathrm{F}=2.304 \times 10^{24} \mathrm{~N}$
Mass of the person, $M=60 \mathrm{~kg}$
Acceleration due to gravity, $\mathrm{g}=9.8 \mathrm{~ms}^{-2}$
Weight $(\mathrm{W})=\mathrm{mg}$
$=60 \times 9.8$
$\mathrm{W}=588 \mathrm{~N}$
Comparison: Electrostatic force is equal to $3.92 \times 10^{21}$ times of weight of the person.
Question 3.
Five identical charges $\mathrm{Q}$ are placed equidistant on a semicircle as shown in the figure. Another point charge $\mathrm{q}$ is kept at the center of the circle of radius $\mathrm{R}$. Calculate the electrostatic force experienced by the charge q.

Solution:
Force acting on $\mathrm{q}$ due to $\mathrm{Q}_1$ and $\mathrm{Q}_5$ are opposite direction, so cancel to each other.
Force acting on $\mathrm{q}$ due to $\mathrm{Q}_3$ is $\mathrm{F}_3=\frac{1}{4 \pi \varepsilon_0} \frac{q Q_3}{R^2}$
Force acting on $\mathrm{q}$ due to $\mathrm{Q}_2$ and $\mathrm{Q}_4$
Resolving in two component method:
(i) Vertical Component:
$\mathrm{Q}_2 \operatorname{Sin} \theta$ and $\mathrm{Q}_4 \operatorname{Sin} \theta$ are equal and opposite direction, so they are cancel to each other.
(ii) Horizontal Component:
$\mathrm{Q}_2 \operatorname{Sin} \theta$ and $\mathrm{Q}_4 \cos \theta$ are equal and same direction, so they can get added.
$F_{24}=F_{2 q}+F_{4 q}=F_2 \cos 55^{\circ}+F_4 \cos 45^{\circ}$
$\mathrm{F}_{24}=\frac{1}{4 \pi \varepsilon_0} \frac{q Q_2}{R^2} \cos 45^{\circ}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q 4}{R^2} \cos 45^{\circ}$
Resultant net force $F$

$
\begin{aligned}
& \mathrm{F}=\mathrm{F}_3+\mathrm{F}_{24}+\mathrm{F}_{15} \\
& \mathrm{Q}=\mathrm{Q}_1=\mathrm{Q}_2=\mathrm{Q}_3=\mathrm{Q}_4=\mathrm{Q}_5 \\
& \cos 45^{\circ}=\frac{1}{\sqrt{2}}
\end{aligned}
$
$
\begin{aligned}
\mathrm{F} & =\frac{q \mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}^2}\left[1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right] \\
& =\frac{1}{4 \pi \varepsilon_0} \frac{q \mathrm{Q}}{\mathrm{R}^2}\left[1+\frac{2}{\sqrt{2}}\right] \\
\mathrm{F} & =\frac{1}{4 \pi \varepsilon_0} \frac{q \mathrm{Q}}{\mathrm{R}^2}[1+\sqrt{2}] \mathrm{N}
\end{aligned}
$
Vector form:
$
\overrightarrow{\mathrm{F}}=\frac{1}{4 \pi \varepsilon_0} \frac{q \mathrm{Q}}{\mathrm{R}^2}(1+\sqrt{2}) \mathrm{N} \hat{i}
$
Question 4.
Suppose a charge $+q$ on Earth's surface and another $+q$ charge is placed on the surface of the Moon, (a) Calculate the value of $q$ required to balance the gravitational attraction between Earth and Moon (b) Suppose the distance between the Moon and Earth is halved, would the charge $q$ change? (Take $\mathrm{m}_{\mathrm{E}}=5.9$ $\times 10^{24} \mathrm{~kg}, \mathrm{~m}_{\mathrm{M}}=7.348 \times 10^{22} \mathrm{~kg}$ )
Solution:
Mass of the Earth, $M_E=5.9 \times 10^{24} \mathrm{~kg}$
Mass of the Moon, $\mathrm{M}_{\mathrm{M}}=7.348 \times 10^{22} \mathrm{~kg}$
Charge placed on the surface of Earth and Moon $=q$
(a) Required charge to balance the $\mathrm{F}_{\mathrm{G}}$ between Earth and Moon
$
\begin{aligned}
& \mathrm{F}_{\mathrm{C}}=\mathrm{F}_{\mathrm{G}} \text { (or) } \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2}=\frac{\mathrm{GM}_{\mathrm{E}} \times \mathrm{M}_{\mathrm{M}}}{r^2} \\
& \mathrm{q}^2=\mathrm{G} \times \mathrm{M}_{\mathrm{E}} \times \mathrm{M}_{\mathrm{M}} \times 4 \pi \varepsilon_0=320.97 \times 10^{25} \\
& \mathrm{q}=\sqrt{320.97 \times 10^{25}}=5.665 \times 10^{13}=5.67 \times 10^{13} \mathrm{C}
\end{aligned}
$

(b) The distance between Moon and Earth is
$
\begin{aligned}
& \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r / 2}=\frac{\mathrm{G} \mathrm{M}_{\mathrm{E}} \times \mathrm{M}_{\mathrm{M}}}{r / 2} \\
& \text { so } \mathrm{q}=5.67 \times 10^{13} \mathrm{C}
\end{aligned}
$
There is no change.
Question 5.
Draw the free body diagram for the following charges as shown in the figure (a), (b) and (c).

Question 6.
Consider an electron travelling with a speed $\mathrm{V}_{\Phi}$ and entering into a uniform electric field $\vec{E}$ which is perpendicular to $\overrightarrow{\mathrm{V}_0}$ as shown in the Figure. Ignoring gravity, obtain the electron's acceleration, velocity and position as functions of time.
Solution:

Speed of an electron $=V_0$
Uniform electric field $=\vec{E}$
(a) Electron's acceleration:
Force on electron due to uniform electric field, $\mathrm{F}=\mathrm{Ee}$
Downward acceleration of electron due to electric field, $\mathrm{a}=\frac{F}{m}=-\frac{e E}{M}$
Vector from, $\vec{a}=-\frac{e E}{M} \hat{j}$
(b) Electron's velocity:
Speed of electron in horizontal direction, $\mathrm{u}=\mathrm{V}_0$ From the equation of motion, $\mathrm{V}=\mathrm{u}+\mathrm{at}$
$\mathrm{V}=\mathrm{V}_0 \frac{e E}{M} \mathrm{t}$
Vector from $\vec{V}=\mathrm{V}_0 \hat{j}-\frac{e E}{M} \mathrm{t} \hat{j}$
(c) Electron's position:
Position of electron, $s=r$
From equation of motion, $r=v_0 t+\frac{1}{2}\left(-\frac{e E}{M}\right) t^2$
$\mathrm{r}=\mathrm{V}_0 \mathrm{t}+\frac{1}{2} \frac{e E}{M} \mathrm{t}^2 \hat{j}$
Vector from,
$
\vec{r}=\mathrm{V}_0 \mathrm{t} \hat{j} \frac{1}{2} \frac{e E}{M} \mathrm{t}^2 \hat{j}
$
Question 7.
A closed triangular box is kept in an electric field of magnitude $\mathrm{E}=2 \times 10^3 \mathrm{~N} \mathrm{C}^{-1}$ as shown in the figure.

Calculate the electric flux through the (a) vertical rectangular surface (b) slanted surface and (c) entire surface.
Answer:
Electric field of magnitude $\mathrm{E}=2 \times 10^3 \mathrm{NC}^{-1}$
(a) Vertical rectangular surface:
Rectangular area $\mathrm{A}=5 \times 10^{-2} \times 15 \times 10^{-2}$
$
\begin{aligned}
& A=75 \times 10^{-24} \mathrm{~m}^2 \\
& \theta=180^{\circ} \\
& \Rightarrow \cos 180^{\circ}=-1
\end{aligned}
$

$
\begin{aligned}
& \text { Electric flux, } \Phi_{\mathrm{v} . \mathrm{s}}=\mathrm{EA} \cos \theta \\
& =2 \times 10^3 \times 75 \times 10^{-4} \times \cos 180^{\circ} \\
& =-150 \times 10^{-1} \\
& \Phi_{\mathrm{v} . \mathrm{s}}=-15 \mathrm{Nm}^2 \mathrm{C}^{-1}
\end{aligned}
$
(b) Slanted surface:
$
\begin{aligned}
& \cos \theta=\cos 60^{\circ}=0.5 \\
& \sin \theta=\sin 30^{\circ}=\frac{\text { Opposite }}{\text { hyp }}
\end{aligned}
$

$
\begin{aligned}
& \text { hyp }=\frac{5 \times 10^2}{0.5} \\
& \text { hyp }=0.1 \mathrm{~m}
\end{aligned}
$
Area of slanted surface $\mathrm{A}_2=\left(0.1 \times 15 \times 10^{-2}\right)$
$
\mathrm{A}_2=0.015 \mathrm{M}^2
$
Electric flux, $\Phi_{\mathrm{v} . \mathrm{s}}=\mathrm{EA}=\cos \theta$
$
\begin{aligned}
& =2 \times 10^3 \times 0.015 \times \cos 60^{\circ} \\
& =2 \times 10^3 \times 0.015 \times 10^3 \\
& =0.015 \times 10^3 \\
& \Phi_{\mathrm{v} . \mathrm{s}}=15 \mathrm{Nm}^2 \mathrm{C}^{-1}
\end{aligned}
$
Horizontal surface
$
\theta=90^{\circ} ; \cos 90^{\circ}=0
$
Electric flux, $\Phi_{\mathrm{H} . \mathrm{S}}=\mathrm{E} \cdot \mathrm{A}_3 \operatorname{Cos} 90^{\circ}=0$
(c) Entire surface:
$\Phi_{\text {Total }}=\Phi_{\mathrm{V} . \mathrm{S}}+\Phi_{\mathrm{S} . \mathrm{S}}+\Phi_{\mathrm{H} . \mathrm{S}}=-15+15+0$
$\Phi_{\text {Total }}=0$
Question 8.
The electrostatic potential is given as a function of $\mathrm{x}$ in figure (a) and (b). Calculate the corresponding electric fields in regions $A, B, C$ and $D$. Plot the electric field as a function of $\mathrm{x}$ for the figure (b).

Answer:
The relation between electric field and potential $\mathrm{E}=-\frac{d v}{d x}$
(a) Region A:
$\mathrm{dv}=-3 \mathrm{~V} ; \mathrm{dx}=0.2 \mathrm{~m}$
Electric field, $\mathrm{E}_{\mathrm{A}}=\frac{(-3)}{0.2}=15 \mathrm{~V} \mathrm{~m}^{-1}$
Region B:
$\mathrm{dv}=0 \mathrm{~V} ; \mathrm{dx}=0.2 \mathrm{~m}$
Electric field, $\mathrm{E}_{\mathrm{B}}=\frac{0}{0.2}=0$
Region C:
$\mathrm{dv}=2 \mathrm{~V} ; \mathrm{dx}=0.2 \mathrm{~m}$
Electric field, $\mathrm{E}_{\mathrm{C}}=\frac{-2}{0.2}=10 \mathrm{~V} \mathrm{~m}^{-1}$
Region D:
$\mathrm{dv}=-6 \mathrm{~V} ; \mathrm{dx}=0.2 \mathrm{~m}$
Electric field, $\mathrm{E}_{\mathrm{D}}=-\left(\frac{-6}{0.2}\right)=10 \mathrm{~V} \mathrm{~m}^{-1}=30 \mathrm{~V} \mathrm{~m}^{-1}$
Electric field, $\mathrm{E}_{\mathrm{A}}=15 \mathrm{~V} \mathrm{~m}^{-1}$
Electric field, $\mathrm{E}_{\mathrm{B}}=0$
Electric field, $\mathrm{E}_{\mathrm{C}}=\frac{(-3)}{0.2}=10 \mathrm{~V} \mathrm{~m}^{-1}$
Electric field, $\mathrm{E}_{\mathrm{D}}=30 \mathrm{~V} \mathrm{~m}^{-1}$

Question 9.
A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around $0.6 \mathrm{~mm}$ gap as shown in the figure.

To create the spark, an electric field of magnitude $3 \times 10^6 \mathrm{Vm}^{-1}$ is required, (a) What potential difference must be applied to produce the spark? (b) If the gap is increased, does the potential difference increase, decrease or remains the same? (c) find the potential difference if the gap is $1 \mathrm{~mm}$.
Answer:
Separation gap between two electrodes, $d=0.6 \mathrm{~mm}$ $\mathrm{d}=0.6 \times 10^{-3} \mathrm{~m}$
Magnetude of electric field Electric field $=\mathrm{E}=3 \times 10^6 \mathrm{~V} \mathrm{~m}^{-1}$
Electric field $\mathrm{E}=\frac{V}{d}$
(a) Applied potential difference, $\mathrm{V}=\mathrm{E} \cdot \mathrm{d}$
$=3 \times 10^6 \times 0.610^{-13}=1.8 \times 10^3$
$\mathrm{V}=1800 \mathrm{~V}$
(b) From equation, $V=E \cdot d$
If the gap (distance) between the electrodes increased, the potential difference also increases.
(c) Gap between the electrodes, $\mathrm{d}=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}$
Potential difference, $V=$ E.d
$
\begin{aligned}
& =3 \times 10^6 \times 1 \times 10^{-3}=3 \times 103 \\
& \mathrm{~V}=3000 \mathrm{~V}
\end{aligned}
$
Question 10.
A point charge of $+10 \mu \mathrm{C}$ is placed at a distance of $20 \mathrm{~cm}$ from another identical point charge of $+10 \mu \mathrm{C}$. A point charge of $-2 \mu \mathrm{C}$ is moved from point $a$ to $\mathrm{b}$ as shown in the figure. Calculate the change in potential energy of the system? Interpret your result.

$
\begin{aligned}
& \mathrm{q}_1=10 \mu \mathrm{C}=10 \times 10^{-6} \mathrm{C} \\
& \mathrm{q}_2=2 \mu \mathrm{C}=-2 \times 10^{-6} \mathrm{C}
\end{aligned}
$
distance, $\mathrm{r}=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$
Answer:
Change in potential energy,
$
\begin{aligned}
& \Delta \mathrm{U}=\frac{9 \times 10^9 \times 10 \times 10^{-6} \times\left(-2 \times 10^{-6}\right)}{\not 6 \times 10^{-2}} \\
& =-36 \times 1 \times 10^9 \times 10^{-12} \times 10^2=-36 \times 10^{-1} \\
& \Delta \mathrm{U}=-3.6 \mathrm{~J}
\end{aligned}
$
Negative sign implies that to move the charge $-2 \mathrm{pC}$ no external work is required. System spends its stored energy to move the charge from point a to point $b$.
Ans:
$\Delta \mathrm{U}=-3.6 \mathrm{~J}$, negative sign implies that to move the charge $-2 \mu \mathrm{C}$ no external work is required. System spends its stored energy to move the charge from point a to point $\mathrm{b}$.
Question 11.
Calculate the resultant capacitances for each of the following combinations of capacitors.

Answer:

Parallel combination of capacitor 1 and 2
$
\mathrm{C}_{\mathrm{p}}=\mathrm{C}_0+\mathrm{C}_0=2 \mathrm{C}_0
$
Series combination of capacitor $C_p$ and 3
$
\frac{1}{C_S}=\frac{1}{C_p}+\frac{1}{C_3}=\frac{1}{2 C_0}+\frac{1}{C_0}=\left(\text { or) } \frac{1}{C_S}=\frac{3}{2} \mathrm{C}_0 \text { (or) } \mathrm{C}_{\mathrm{S}}=\frac{2}{3} \mathrm{C}_0\right.
$

$
\begin{aligned}
& \frac{1}{C_{S_1}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{C_0}+\frac{1}{C_0}=\frac{1}{C_0} \text { (or) } \\
& \frac{1}{C_{S_1}}=\frac{2}{C_0} \text { (or) } C_{S_1}=\frac{C_0}{2}
\end{aligned}
$
Similarly 3 and 4 are series combination
$
\frac{1}{C_{S_2}}=\frac{1}{C_3}+\frac{1}{C_4}=\frac{1}{C_0}+\frac{1}{C_0}=\frac{2}{C_0} \text { (or) } C_{S_2}=\frac{C 0}{2}
$

$C_{S_1}$ and $C_{S_2}$ are in parallel combination
$
\mathrm{C}_{\mathrm{p}}=C_{S_1}+C_{S_2}=\frac{C_0}{2}+\frac{C_0}{2} \text { (or) } \mathrm{C}_{\mathrm{p}}=\frac{2 C_0}{2} \mathrm{C}_{\mathrm{p}}=\mathrm{C}_0
$
(c) Capacitor 1,2 and 3 are in parallel combination
$
\begin{aligned}
& \mathrm{C}_{\mathrm{p}}=\mathrm{C}_0+\mathrm{C}_0+\mathrm{C}_0=3 \mathrm{C}_0 \\
& \mathrm{C}_{\mathrm{p}}=3 \mathrm{C}_0
\end{aligned}
$

(d) Capacitar $\mathrm{C}_1$ and $\mathrm{C}_2$ are in combination
$
\begin{gathered}
\frac{1}{\mathrm{C}_{\mathrm{S}_1}}=\frac{\mathrm{C}_1+\mathrm{C}_2}{\mathrm{C}_1 \mathrm{C}_2} \\
\mathrm{C}_{\mathrm{S}_1}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}
\end{gathered}
$
Similarly $\mathrm{C}_3$ and $\mathrm{C}_4$ are in series combination

$
\begin{aligned}
& \frac{1}{\mathrm{C}_{\mathrm{S}_2}}=\frac{1}{\mathrm{C}_3}+\frac{1}{\mathrm{C}_4}=\frac{\mathrm{C}_3+\mathrm{C}_4}{\mathrm{C}_3 \mathrm{C}_4} \\
& \mathrm{C}_{\mathrm{S}_2}=\frac{\mathrm{C}_3 \mathrm{C}_4}{\mathrm{C}_3+\mathrm{C}_4}
\end{aligned}
$
$C_{S_1}$ and $C_{S_2}$ are in parallel combination across RS:

$\begin{aligned}
& C P=C_{S_1}+C_{S_2} \\
& =\frac{C_1 C_2}{C_1+C_2}+\frac{C_3 C_4}{C_3+C_4}=\frac{C_1 C_2\left(C_3+C_4\right)+C_3 C_4\left(C_1+C_2\right)}{\left(C_1+C_2\right)\left(C_3+C_4\right)} \\
& C_P=\frac{C_1 C_2 C_3+C_1 C_2 C_4+C_3 C_4 C_1+C_3 C_4 C_2}{\left(C_1+C_2\right)\left(C_3+C_4\right)}
\end{aligned}$

(e) Capacitor 1 and 2 are series combination

$
\begin{aligned}
& \frac{1}{\mathrm{C}_{\mathrm{S}_1}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}=\frac{1}{\mathrm{C}_0}+\frac{1}{\mathrm{C}_0}=\frac{1}{\mathrm{C}_0} \\
& \frac{1}{\mathrm{C}_{\mathrm{S}_1}}=\frac{2}{\mathrm{C}_0} \text { (or) } \mathrm{C}_{\mathrm{S}_1}=\frac{\mathrm{C}_0}{2}
\end{aligned}
$
Similarly 3 and 4 are series combination
$
\frac{1}{C_{S_2}}=\frac{2}{C_0} \text { (or) } C_{S_2}=\frac{C_0}{2}
$
Three capacitors are in parallel combination

$
\begin{aligned}
& \mathrm{C}_{\mathrm{P}}=\frac{\mathrm{C}_0}{2}+\mathrm{C}_0+\frac{\mathrm{C}_0}{2}=\frac{\mathrm{C}_0}{2}+\frac{2 \mathrm{C}_0}{2}+\frac{\mathrm{C}_0}{2}=\frac{4 \mathrm{C}_0}{2} \\
& \mathrm{C}_{\mathrm{P}}=2 \mathrm{C}_0
\end{aligned}
$
Question 12
An electron and a proton are allowed to fall through the separation between the plates of a parallel plate capacitor of voltage $5 \mathrm{~V}$ and separation distance $\mathrm{h}=1 \mathrm{~mm}$ as shown in the figure.

(a) Calculate the time of flight for both electron and proton
(b) Suppose if a neutron is allowed to fall, what is the time of flight?
(c) Among the three, which one will reach the bottom first?
(Take $\mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}, \mathrm{~m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}$ and $\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}$ )
Answer:
Potential difference between the parallel plates $\mathrm{V}=5 \mathrm{~V}$
Separation distance, $\mathrm{h}=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}$
Mass of proton, $\mathrm{mp}=1.6 \times 10^{-27} \mathrm{~kg}$
Mass of proton, $\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$
Charge of an a proton (or) electron, e $-1.6 \times 10^{-19} \mathrm{C}$
$[\mathrm{u}=0 ; \mathrm{s}=\mathrm{h}]$
From equation of motion, $S=u t+\frac{1}{2} a^2$
From equation of motion, $\mathrm{h}=\frac{1}{2} \mathrm{at}^2$
$
\mathrm{t}=\sqrt{\frac{2 h}{a}}
$
Acceleration of an electron due to electric field, $\mathrm{a}=\frac{F}{m}=\frac{e E}{m}$ $\left[\mathrm{E}=\frac{V}{d}\right]$
(a) Time of flight for both electron and proton,
$
\begin{aligned}
t_e & =\sqrt{\frac{2 h m_e}{e \mathrm{E}}}=\sqrt{\frac{2 \times 1 \times 10^{-13} \times 9.1 \times 10^{-31} \times 10^{-3}}{1.6 \times 10^{-19} \times 5}} \\
& =\sqrt{\frac{18.2 \times 10^{-34} \times 10^{-3}}{8 \times 10^{-19}}}=\sqrt{2.275 \times 10^{-15} \times 10^{-3}} \\
t_e & =1.5 \times 10^{-9} \mathrm{~s} \\
t_e & =1.5 \mathrm{~ns} \\
t_p & =\sqrt{\frac{2 \mathrm{hm}_e}{e \mathrm{E}}}=\sqrt{\frac{2 h m_p \cdot d}{e \mathrm{~V}}}=\sqrt{\frac{2 \times 1 \times 10^{-3} \times 1.6 \times 10^{-27} \times 10^{-3}}{1.6 \times 10^{-19} \times 5}} \\
& =\sqrt{\frac{2 \times 10^{-33}}{5 \times 10^{-19}}}=\sqrt{0.4 \times 10^{-14}}=6.32 \times 10^{-8} \\
t_p & =63 \times 10^{-9} \\
\mathrm{t}_{\mathrm{p}} & =63 \mathrm{~ns} \ldots \ldots .(2)
\end{aligned}
$

(b) time of flight of neutron $\mathrm{t}_{\mathrm{n}}=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 1 \times 10^3}{10}}=\sqrt{0.2 \times 10^{-3}}$
$
\begin{aligned}
& \mathrm{t}_{\mathrm{n}}=0.0141 \mathrm{~s}=14.1 \times 10^{-3} \mathrm{~s} \\
& \mathrm{t}_{\mathrm{n}}=14.1 \times 10^{-3} \mathrm{~ms} \ldots . . \text { (3) }
\end{aligned}
$
(c) Compairision of values 1,2 and 3 . The electron will reach the bottom first.
Question 13.
During a thunder storm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air $\left(3 \times 10^6 \mathrm{Vm}^{-1}\right)$, lightning will occur.

(a) If the bottom part of the cloud is $1000 \mathrm{~m}$ above the ground, determine the electric potential difference that exists between the cloud and ground.
(b) In a typical lightning phenomenon, around $25 \mathrm{C}$ of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?
Answer:
(a) Electric field between the cloud and ground,
$\mathrm{V}=\mathrm{E} \cdot \mathrm{d}$
$\mathrm{V}=3 \times 10^6 \times 1000=3 \times 10^9 \mathrm{~V}$
(a) Electrons transfered from cloud to ground, $\mathrm{q}=25 \mathrm{C}$
Electron static potential energy,
$
\begin{aligned}
& \mathrm{U}=\frac{1}{2} \mathrm{CV}^2 \\
& {\left[\mathrm{C}=\frac{q}{V}\right]} \\
& =\frac{1}{2} \mathrm{qV}=\frac{1}{2} \times 25 \times 3 \times 10^9 \\
& \mathrm{U}=37.5 \times 10^9 \mathrm{~J}
\end{aligned}
$
Question 14.
For the given capacitor configuration
(a) Find the charges on each capacitor
(b) potential difference across them
(c) energy stored in each capacitor.

Answer:
Capacitor $\mathrm{b}$ and $\mathrm{c}$ in parallel combination
$
\mathrm{C}_{\mathrm{p}}=\mathrm{C}_{\mathrm{b}}+\mathrm{C}_{\mathrm{c}}=(6+2) \mu \mathrm{F}=8 \mu \mathrm{F}
$
Capacitor $\mathrm{a}, \mathrm{c}_{\mathrm{p}}$ and $\mathrm{d}$ are in series combination, so the resulatant copacitance
$
\begin{aligned}
& \frac{1}{C_s}=\frac{1}{C_a}+\frac{1}{C_{c p}}+\frac{1}{C_d}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8} \\
& \mathrm{C}_{\mathrm{S}}=\frac{8}{3} \mu \mathrm{F}
\end{aligned}
$

(a) Charge on each capacitor,
Charge on capacitor $\mathrm{a}, \mathrm{Q}_{\mathrm{a}}=\mathrm{C}_{\mathrm{s}} \mathrm{V}=\frac{8}{3} \times 9$
$
\mathrm{Q}_{\mathrm{a}}=24 \mu \mathrm{C}
$
Charge on capacitor, $d, Q_d=C_{\mathrm{s}} V=\frac{8}{3} \times 9$
$
\mathrm{Q}_{\mathrm{d}}=24 \mu \mathrm{C}
$
Capacitor $b$ and $c$ in parallel
Charge on capacitor, $\mathrm{b}, \mathrm{Q}_{\mathrm{b}}=\frac{6}{3} \times 9=18$
$
\mathrm{Q}_{\mathrm{b}}=18 \mu \mathrm{C}
$
Charge on capacitor, c, $Q_c=\frac{2}{3} \times 9=6$
$
\mathrm{Q}_{\mathrm{c}}=6 \mu \mathrm{C}
$
(b) Potential difference across each capacitor, $\mathrm{V}=\frac{q}{C}$
Capacitor $\mathrm{C}_{\mathrm{a}}, \mathrm{V}_{\mathrm{a}}=\frac{q_a}{C_a}=\frac{24 \times 10^6}{8 \times 10^6}=3 \mathrm{~V}$
Capacitor $\mathrm{C}_{\mathrm{b}}, \mathrm{V}_{\mathrm{b}}=\frac{q_{\mathrm{b}}}{C_b}=\frac{18 \times 10^6}{6 \times 10^6}=3 \mathrm{~V}$
Capacitor $\mathrm{C}_{\mathrm{C}}, \mathrm{V}_{\mathrm{c}}=\frac{q_c}{C_c}=\frac{6 \times 10^6}{2 \times 10^6}=3 \mathrm{~V}$
Capacitor $\mathrm{C}_{\mathrm{d}}, \mathrm{V}_{\mathrm{d}}=\frac{q_d}{C_d}=\frac{24 \times 10^6}{8 \times 10^6}=3 \mathrm{~V}$
(c) Energy stores in a capacitor, $\mathrm{U}=\frac{1}{2} \mathrm{CV}^2$
Energy in capacitor $\mathrm{C}_{\mathrm{a}}, \mathrm{U}_{\mathrm{a}}=\frac{1}{2} \mathrm{C}_{\mathrm{a}} V_a^2=\frac{1}{2} \times 8 \times 10^{-6} \times(3)^2$
$
\mathrm{U}_{\mathrm{a}}=36 \mu \mathrm{J}
$
Capacitor $\mathrm{C}_{\mathrm{b}}, \mathrm{U}_{\mathrm{b}}=\frac{1}{2} \mathrm{C}_{\mathrm{b}} V_b^2=\frac{1}{2} \times 6 \times 10^{-6} \times(3)^2$
$\mathrm{U}_{\mathrm{a}}=27 \mu \mathrm{J}$
$\mathrm{C}_{\mathrm{c}}, \mathrm{U}_{\mathrm{c}}=\frac{1}{2} \mathrm{C}_{\mathrm{c}} V_c^2=\frac{1}{2} \times 2 \times 10^{-6} \times(3)^2$
$\mathrm{U}_{\mathrm{a}}=9 \mu \mathrm{J}$
$\mathrm{C}_{\mathrm{d}}, \mathrm{U}_{\mathrm{d}}=\frac{1}{2} \mathrm{C}_{\mathrm{d}} V_d^2=\frac{1}{2} \times 8 \times 10^{-6} \times(3)^2$
$\mathrm{U}_{\mathrm{a}}=36 \mu \mathrm{J}$
Question 15.
Capacitors $P$ and $Q$ have identical cross sectional areas $A$ and separation $\mathrm{d}$. The space between the capacitors is filled with a dielectric of dielectric constant er as shown in the figure. Calculate the capacitance of capacitors $P$ and $Q$.

Answer:
Cross-sectional area of parallel plate capacitor $=\mathrm{A}$
Each area of different medium between parallel plate capacitor $=\frac{A}{2}$
Separation distance $=\mathrm{d}$
Capacitance of parallel plate capacitor, $\mathrm{C}=\frac{\varepsilon A}{d}$
Air medium of dielectric constant, $\varepsilon_{\mathrm{r}}=1$
dielectric medium of dielectric constant $=\varepsilon_{\mathrm{T}}$
Case 1:
Capacitance of air filled capacitor
$
\mathrm{C}_a=\frac{\varepsilon_0 \varepsilon_r(\mathrm{~A} / 2)}{d}=\frac{\varepsilon_0(1) \mathrm{A}}{2 d}=\frac{\varepsilon_0 \mathrm{~A}}{2 d}
$
Capacitance of dielectric filled capacitor
$
\mathrm{C}_d=\frac{\varepsilon_0 \varepsilon_r(\mathrm{~A} / 2)}{d}=\frac{\varepsilon_0 \varepsilon_r \mathrm{~A}}{2 d}
$
Capacitance of parallel plate capacitor
$
\begin{aligned}
& \mathrm{P}, \mathrm{C}=\mathrm{C}_a+\mathrm{C}_d=\frac{\varepsilon_0 \mathrm{~A}}{2 d}+\frac{\varepsilon_0 \varepsilon_r \mathrm{~A}}{2 d} \\
& \mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{2 d}\left(1+\varepsilon_r\right)
\end{aligned}
$

Case $2:$
Each distance of different medium between the parallel plate capacitor $=\frac{d}{2}$ Capacitance of dielectric filled capacitor
$
\mathrm{C}_d=\frac{\varepsilon_0 \varepsilon_r \mathrm{~A}}{d / 2}=\frac{2 \varepsilon_0 \varepsilon_r \mathrm{~A}}{d}
$
Capacitance of air filled capacitor,
$
\mathrm{C}_a=\frac{\varepsilon_0 \varepsilon_r \mathrm{~A}}{(d / 2)}=\frac{2 \varepsilon_0(1) \mathrm{A}}{d}=\frac{2 \varepsilon_0 \mathrm{~A}}{d}
$
Capacitance of parallel plate capacitor,
$
\begin{aligned}
& \mathrm{Q}=\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_d}+\frac{1}{\mathrm{C}_a}=\frac{1}{\left(\frac{2 \varepsilon_0 \varepsilon_r \mathrm{~A}}{d}\right)}+\frac{1}{\left(\frac{2 \varepsilon_0 \mathrm{~A}}{d}\right)} \\
& \frac{1}{\mathrm{C}}=\frac{d}{2 \varepsilon_0 \varepsilon_r \mathrm{~A}}+\frac{d}{2 \varepsilon_0 \mathrm{~A}}=\frac{d}{2 \varepsilon_0 \mathrm{~A}}\left(\frac{1}{\varepsilon_r}+1\right)=\frac{d}{2 \varepsilon_0 \mathrm{~A}}\left(\frac{1+\varepsilon_r}{\varepsilon_r}\right) \\
& \mathrm{C}=\frac{2 \varepsilon_0 \mathrm{~A}}{d}\left(\frac{\varepsilon_r}{1+\varepsilon_r}\right)
\end{aligned}
$