2 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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Question 12 Marks

Drive the equation of Energy stored in capacitor $\left( U =\frac{ Q ^2}{2 C }\right)$

Answer

→To find the energy stored in capacitor, suppose the charge on the conductor is zero, initially. →Suppose the positive charge is taken from conductor-2 to conductor 1 , bit by bit. At the end of the process, suppose the conductor 1 gets charge Q and conductor 2 has charge - Q .
→To transfer the positive charge from conductor 2 to conductor 1 , work needs to be done and the energy equivalent to this work is stored in the capacitor, which is known as the energy stored in the capacitor.
→Consider the intermediate situation when the conductors 1 and 2 have charges $Q^{\prime}$ and $-Q^{\prime}$ respectively. At this stage, the p.d. $V ^{\prime}$ between the conductors 1 and 2 is $\frac{Q^{\prime}}{C}$ where C is the capacitance of the system.
→Now, work done to transfer a small charge $\delta Q^{\prime}$ from conductor 2 to conductor 1 ,
$\delta W = V ^{\prime} \delta Q ^{\prime}=\frac{ Q ^{\prime}}{ C } \delta Q ^{\prime}$
→Total work required to be done in building (/ establishing) + Q charge on conductor 1 ,
$\begin{aligned}
W & =\int_0^{ Q } \frac{ Q ^{\prime}}{ C } \delta Q ^{\prime} \\
\therefore W & =\frac{1}{ C }\left(\frac{ Q ^{\prime 2}}{2}\right)_0^{ Q }\left[\because \int x^n d x=\frac{x^{n+1}}{n+1}\right] \\
\therefore W & =\frac{1}{ C }\left(\frac{ Q ^2}{2}-0\right) \\
\therefore W & =\frac{ Q ^2}{2 C }
\end{aligned}$
→Because this work is stored in capacitor in the form of energy which is called energy stored in capacitor.
$\therefore U =\frac{ Q ^2}{2 C }$
→Using $Q = CV$ in above equation, we can easily get other alternate forms of above equation.
→$U =\frac{ Q ^2}{2 C }=\frac{1}{2} CV ^2=\frac{1}{2} VQ$

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Question 22 Marks

Explain the effect of dielectric on the capacitance of parallel plate capacitor and derive the equation of dielectric constant.

Answer

$\rightarrow$ Suppose, for a parallel plate capacitor, the separation between plates is $d$ , area of each plate is $A ,$ charge on plate is $\pm Q$ and surface charge density is $\pm \sigma$.
$\rightarrow$ When there is vacuum between two plates, the electric field between them will be
$E _0=\frac{\sigma}{\varepsilon_0}$
and the $p.d.$ will be $V _0= E _0 d$
$\rightarrow$ In this case, capacitance,
$C _0=\frac{ Q }{ V _0}=\frac{\varepsilon_0 A}{d}$
$\rightarrow$ Now, suppose the region between the two plates is totally filled with dielectric material.
$\rightarrow$ Consequently, electric charge is induced on the surface of the dielectric. Surface charge density of this charge is $\pm \sigma_{ P }$.
$\rightarrow$ Because of which electric field is created inside the dielectric $\left(\frac{\sigma_{ P }}{\varepsilon_0}\right)$ which is in direction opposite to the electric field generated by the plates.
$\rightarrow$ Net electric field between two plates of the capacitor,
$E =\frac{\sigma-\sigma_{ P }}{\varepsilon_0}$
$\rightarrow\ p.d.$ between two plates will be
$V = E d$
$ =\left(\frac{\sigma-\sigma_P}{\varepsilon_0}\right) \cdot d$
$\rightarrow$ For a linear dielectric $\sigma_p$ is proportional to $E _0$, which means it is proportional to $\sigma\left(\because E _0=\frac{\sigma}{\varepsilon_0}\right)$
$\therefore \sigma-\sigma_{ P } \alpha \sigma$
$\therefore \sigma-\sigma_{ P }=\frac{\sigma}{ K }$
$\rightarrow$ Here, $K$ is proportionality constant, which is called dielectric constant.
$($For insulators, $K >1)$
$\rightarrow$ Substituting this value in the above equation of $V$,
$\therefore V =\frac{\sigma d}{\varepsilon_0 K}=\frac{ Q d}{\varepsilon_0 AK } \text { (where, } \sigma=\frac{ Q }{ A } \text { ) } \ldots$
$\rightarrow$ When there is dielectric between two plates of capacitor, capacitance
$C =\frac{ Q }{ V }=\frac{\varepsilon_0 KA }{d}$
$\rightarrow \varepsilon_0 K$ is called the permittivity of given medium, which is also denoted by $\varepsilon$.
$\therefore \varepsilon=\varepsilon_0 K$
$K=\frac{\varepsilon}{\varepsilon_0} ($For vacuum, $K=1 \therefore \varepsilon=\varepsilon_0 ) ...$
$\frac{\varepsilon}{\varepsilon_0}$ is also denoted by $\varepsilon_{ r }$
which is known as relative permittivity.
$\rightarrow$ Taking the ratio of $C$ and $C _0$,
$\frac{ C }{ C _0}= K($ from eq. $(1)$ and $(6))$
$\therefore C = K C _0$

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Question 32 Marks

Mention two important conclusions regarding the relation between electric field and electric potential.

Answer

→(i) Electric field is in the direction in which the electric potential decreases steepest.
(ii) The magnitude of electric field is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

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Question 42 Marks

Write a note on equipotential surface.

Answer

"An imaginery surface, any surface that have same electric potential at every point on it is called equipotential surface."

$\rightarrow $ Electric potential at distance $r$ from a point charge $q$,
$V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \text { If } r \text { is constant then }$
$V =\text { const. also }$
$\rightarrow $ Hence, equipotential surface for a point charge $q$ are different spheres having charge $q$ at centre, and having different radii. $($Fig. a$)$
$\rightarrow $ For such two different equipotential surfaces, potential will be different but potential for all points on any one equipotential surface will be same.
$\rightarrow $ Electric field lines for a point charge are radial lines starting $($originating$)$ from the charge and ending at infinity. $($or in negative charge.$)$
So we can say that electric field lines are perpendicular to equipotential surface.
$\rightarrow $ If the electric field is not perpendicular to the equipotential surface then there will be a component of $\overrightarrow{ E } ($electric field$)$ parallel to the surface. Which shows work needs to be done in moving the unit test charge against this component of electric field, which is against the definition of equipotential surface.
$\rightarrow $ Electric potential difference between any two points on the equipotential surface is zero. $( V = 0 )$
so, no work is required to be done in moving test charge on the surface.
$\rightarrow $ Hence electric field at each point on the equipotential surface must be perpendicular to the surface.
$\rightarrow $ Equipotential surfaces for uniform electric field are shown in fig. $C .$
$\rightarrow $ If the field is along $X-$direction, equipotential surface must be parallel to $YZ -$ plane.

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Question 52 Marks

Write down the equation of electric potential for a point $-$ outside the spherical shell $-$ on the surface of a spherical shell and $-$ inside the spherical shell.

Answer

$\rightarrow $ Generally, for a spherical shell, the electric field for a point outside the surface of shell is such that the entire charge of shell can be considered as concentrated at its centre.
$\rightarrow $ The value of electric potential that we get for a point outside the surface of shell is also got $($derived$)$ in a similar way.
$V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}(r \geq R )$
Where, $q -$ total electric charge on the shell
$R -$ radius of the shell
$r -$ the distance of the point from the centre of the shell.
$\rightarrow $ The electric field inside the shell is zero, therefore potential inside remains constant.
$\rightarrow $ Because as there is no electric field inside, the work done in taking the charge from one point to the other is zero.
$\therefore \text { In } W =q \Delta V ,$
$\text { because } W =0$
$\therefore \Delta V =0$
$\therefore V =\text { constant }$
and that value of potential is same as the value on the surface of shell.
$\therefore V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{ R } \quad(\because r \leq R )$
$\rightarrow $ Below figure $($graph$)$ shows the variation of electric potential $( V )$ versus distance from the

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Question 62 Marks

Draw the graph of variation in potential difference and electric field with distance ' $r$ ' for a point electric charge $Q$.

Answer

Electric potential of a point charge,
$V =\frac{k Q }{r} \text { and }$

electric field $E =\frac{k Q }{r^2}$
$\therefore V \propto \frac{1}{r}$ and
$\text { E } \alpha \frac{1}{r^2}$

In the figure(graph), how electric potential and electric field changes with distance $r$ is shown.

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Question 72 Marks

Show that the interior of a conductor can have no excess charge in the static situation.

Answer

→A neutral conductor has equal amounts of positive and negative charges in every small volume or surface element. When the conductor is charged, the excess charge can reside only on the surface in the static situation.
→Consider an arbitrary volume element V inside a conductor. On the closed surface ' $S$ ' bounding (enclosing) the volume element V , electric field is zero. Hence, by Gauss's law, there is no net charge enclosed by S .
→This means there is no net charge at any point inside the conductor and any excess charge must reside at the surface.

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Question 82 Marks

Show that the electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface.

Answer

→Electric field inside a conductor is zero. Therefore, there is no tangential component of $\overrightarrow{ E }$ on the surface of conductor. No work is done for moving a small test charge within the conductor and on its surface. There is no p.d. between any two points inside or on the surface of the conductor.
→If the conductor is charged, electric field is normal to the surface. This means potential will be different for the surface and a point just outside the surface.
→In a system of conductors of arbitrary size, shape and charge distribution (configuration), each conductor is characterized by a constant value of potential but this constant may differ from one conductor to the other.

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Question 92 Marks

Derive the equation $($formula$)$ showing the relationship between electric field and electric potential.

Answer

$\rightarrow$ As shown in the fig., two equipotential surfaces $A$ and $B$ are very close to each other.
Magnitudes of electric potentials on them are $V$ and $V +\delta V$ respectively.
$\rightarrow$ Here, $\delta V$ is change in electric potential in the direction of electric field $\vec{E}$.
$\rightarrow$ Point $P$ is present on surface $B$.
And the perpendicular distance from surface $A$ to point $P$ is $\delta l$.
$\rightarrow $ The amount of work done in taking a unit positive charge on the perpendicular line from surface $B$ to surface $A$ is equal to $|\vec{E}| \delta l$.
This work is equal to the electric potential difference between surfaces $A$ and $B$, which is $V_A-V_B$.
$\therefore|\overrightarrow{ E }| \delta l=\Delta V = V _{ A }- V _{ B }$
$\therefore|\overrightarrow{ E }| \cdot \delta l= V -( V +\delta V )$
$=-\delta V$
$\therefore \quad|\vec{E}|=-\frac{\delta V }{\delta l}$
$\rightarrow$ Here, $\delta V$ is negative, so taking $-\delta V$ in place of $\delta V$,
$|\overrightarrow{ E }|=\frac{\delta V }{\delta l}$

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Question 102 Marks

On which two factors the extent of polarization in a polar dielectric depends ? what is polarization of dielectric ? Write its formula.

Answer

$\rightarrow$ The extent of polarization in a dielectric depends on the relative strenghts of two mutually opposing factors :
$(i)$ The dipole potential energy in external field tending to align the dipoles with the field.
$(ii)$ The thermal energy tending to disrupt the alignment.
$\rightarrow$ Thus, in either case, whether polar or non $-$ polar, a dielectric develops a net dipole moment in the presence of an external field.
$\rightarrow$ "The net dipole moment per unit volume is called polarization." and is denoted by $\vec{P}$.
$\rightarrow$ For linear isotropic dielectrics, the polarization is found to be proportional to the applied electric field $\vec{E}$.
$\therefore \overrightarrow{ P } \alpha \overrightarrow{ E }$
$\therefore \overrightarrow{ P }=\left(\varepsilon_0 \chi_e\right) \overrightarrow{ E }$
where $\chi_e-$ is a characteristic constant of dielectric and is known as the electric suceptibility of the given dielectric medium.
and $\vec{E}=$ External electric field.
Unit of $\chi_e$ is : unitless and dimension less.

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Question 112 Marks

Derive the equation of P.E. (Potential Energy) of an electric dipole in uniform electric field.

Answer

→Consider an electric dipole as shown in figure at an angle $\theta$ in the external electric field.
→ In the uniform electric field, the net force acting on dipole is zero, but net torque acts on the dipole, which is given by formula :
$\begin{aligned}
\vec{\tau} & =\vec{p} \times \vec{E} \\
\therefore \quad \tau & = PE \sin \theta
\end{aligned}$

(where, $\theta$ is the angle between $\vec{p}$ and $\overrightarrow{ E }$.)
→When electric dipole moment $\vec{p}$ and electric field $\vec{E}$ are not parallel or antiparallel in that condition this torque tends to rotate the electric dipole.
→Suppose, an external torque $\tau_{\text {ext }}$ is applied on the electric dipole is such a way that it cancel this torque.

→Now, the work done by the external torque in rotating the dipole from angle $\theta_0$ to $\theta$, without angular acceleration,
$\begin{aligned}
W & =\int_{\theta_0}^{\theta_1} \tau_{\text {ext }} d \theta \\
\therefore \quad W & =\int_{\theta_0}^{\theta_1} p E \sin \theta d \theta \\
\therefore W & =p E (-\cos \theta)_{\theta_0}^{\theta_1} \\
\therefore \quad W & =p E \left(\cos \theta_0-\cos \theta_1\right)
\end{aligned}$

→This work is stored in the form of potential energy of dipole.

→Therefore P.E. of dipole,
$U =p E \left(\cos \theta_0-\cos \theta_1\right)$

→If the dipole is perpendicular to the electric field initially,
Taking $\theta_0=\frac{\pi}{2}$,
$\cos \theta_0=\cos \frac{\pi}{2}=0 .$
→P.E. of dipole $= U =-p E \cos \theta_1$
$\therefore U =-\vec{p} \cdot \vec{E}$

Where, $\theta_1$ is the angle between $\overrightarrow{ P }$ and $\overrightarrow{ E }$.

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Question 122 Marks

Provide a brief information about dielectric materials.

Answer

→Non-conducting materials are called dielectrics.
→Dielectrics are of 2 types :
(1) Polar dielectric
(2) Non-polar dielectric
(1) Polar dielectric :
→The type of dielectric material in which the centres of a positive and a negative electric charge do not coincide is called a polar dielectric.
→The molecules of a polar dielectric possess permanent electric dipole moment.
Example : $H _2 O , HCl$, etc.

(2) Non-polar dielectric :
→The dielectric in which the centres of positive and negative charges coincide is called nonpolar dielectric.

→The molecules of a non-polar dielectric do not possess permanent dipole moment. Exa., $H _2, O _2, N_2$ etc.

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Question 132 Marks

Explain dielectric strength and dielectric breakdown.

Answer

→"An insulating medium which retains its insulating property up to a maximum electric field is called the dielectric strength of the insulating medium."
OR
→"The minimum electric field which starts the ionization a given non-conducting medium is called its dielectric strength."
→ A dielectric medium loses its insulating property many times, due to applied electric field. This effect is called dielectric breakdown.
→ For air, dielectric strength is nearly $3 \times 10^6 V / m$. (or $3000 V / mm$ )

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Question 142 Marks

Explain the difference between behaviour of a conductor and a dielectric inside external electric field.

Answer

• Conductor in external electric field :
→ As shown in figure (a) when a conductor is placed in external electric field, static electric charge is induced on its surface. Electric field is created inside the conductor due to this electric charge.
→The electric field generated due to induced charges, opposes the external electric field in the region inside the conductor.
→In stable situation, both these electric fields cancel out their mutual effect. When both these electric fields are of equal magnitude the net electric field inside conductor vanishes to zero.
• Dielectric in the external electric field :
→Non-conducting material is called dielectric. Free motion of electric charges is not possible in dielectric.
→The external electric field induces dipole moment either by pulling the molecules of dielectric or by their re-arrangement. Due to which some net charge is induced on the surface of the dielectric.
→The electric field produced by this electric charge opposes the external electric field.
→Due to which electric field inside the dielectric decreases.
→In non-conducting materials, magnitudes of both these electric fields cannot be equal and hence net electric field inside die-electric is not zero.

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Question 152 Marks

Derive the formula of electric potential energy of system of two charges.

Answer

$\rightarrow$ Suppose, two charges $q_1$ and $q_2$ are at initially at infinity.

$\rightarrow$ In the absence of external electric field work done in bringing charge $q_1$ from infinity to a point having position vector $\overrightarrow{r_1}$ is zero.
$\rightarrow$ Electric potential in the space at point $P$ due to charge $q_1$
$V _1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_{1 p}}$
Where, $r_{1 p}=$ distance from charge $q_1$ to point $P$ .
$\rightarrow$ From the definition of electric potential.
In the presence of electric field, work done in brining
charge $q_2$ from infinity to the point having position vector $\overrightarrow{r_2}$,
$W =\binom{\text { Charge }}{q_2} \times\binom{\text { Electric potential at }}{\text { distance } r_{12} \text { due to } q_1}$
$W=q_2\left(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_{12}}\right)$
Where, $r_{12}=$ distance between point $1$ and $2$
$\therefore W =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r_{12}}$
$\rightarrow$ This work is stored in the form of potential energy of the system.
So, the potential energy of the system of the two charges,
$U =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r_{12}}$
$\rightarrow$ This equation is true for any signs of charges $q_1$ and $q_2 \ ($positive as well as negative charges$).$
$\rightarrow$ If $q_1 q_2>0$, the potential energy is positive.
which means work done on the electric charge is positive.
$\rightarrow$ If $q_1 q_2<0$, the potential energy is negative, which means the work required to be done on electric charge is negative.

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Question 162 Marks

Write the equation of electric potential for electric dipole and discuss its special cases.

Answer

$\rightarrow$ The formula of electric potential of electric dipole is as follows :
$V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{p \cos \theta}{r^2}$
OR
$V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\vec{p} \cdot \vec{r}}{r^3}$
Where; $p=$ Electric dipole moment
$\theta=$ angle between position vector $\vec{r}$ and dipole moment $\vec{p}$.
$•$ Special cases :
$(i)$ The point at which electric potential is to be derived, is on the axis of the dipole.
$\therefore \theta=0$
OR
$\theta=\pi$
$\therefore V = \pm \frac{1}{4 \pi \varepsilon_0} \cdot \frac{p}{r^2}$
$ (\because \cos \theta=1 \text { or } \cos \theta=-1)$
$(ii)$ The point at which electric potential is to be found $($/derived$)$ is on equatorial axis of dipole.
$\therefore \theta=\frac{\pi}{2}$
$\therefore \cos \theta=\cos \frac{\pi}{2}=0$
$\therefore V =0$
So, potential on the equatorial axis of dipole is zero.

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Question 172 Marks

Explain $($or write a short note on$)$ capacitors and capacitance.

Answer

$\rightarrow$ A system of two conductors separated by a dielectric medium, is called a capacitor.
$\rightarrow$ As shown in fig., two isolated/ or separated) conductors have charges $+Q$ and $-Q$ and their potentials are $V_1$ and $V_2$ respectively, and $p.d.$ between them is $V=V_1-V_2$.
$\rightarrow$ These conductors can be charged by connecting them to a battery. Charge on any one of the two conductors is called the charge of capacitor.
$\rightarrow$ Total $($Net$)$ charge of the capacitor is zero.
$\rightarrow$ The electric field in the region between the conductors is proportional to charge $Q . ( E \propto Q )$
$\rightarrow$ The p.d. between two conductors means the work done per unit positive charge in taking a small test charge from the conductor $2$ to conductor $1$ against electric field. As the charge increases, potential also increases.
$\rightarrow$ Consequently, $V \propto Q$
and therefore, the ratio $\frac{ Q }{ V }$ is constant.
$\therefore C =\frac{ Q }{ V }$
where constant C is called the capacitance of the capacitor.
$\rightarrow$ C is independent of $Q$ and $V .$
$\rightarrow$ C depends on geometric configuration $($shape, size, separation$)$ of the system of two conductors, and on the dielectric constant of the material placed between two conductors.
$\rightarrow$ SI unit of capacitance is $F ($farad$) OR C/V 1 F=1 C$. $V ^1$
$\rightarrow$ Its dimensional formula is $M ^{-1} L^{-2} T^4 A^2$
$\rightarrow$ Smaller units used in day to day life :
$1 \mu F =10^{-6} F 1 p F=10^{-12} F$
$1 n F=10^{-9} F$
$\rightarrow$ The circuit symbol of a capacitor with fixed capacitance is :

$\rightarrow$ While the capacitor with variable capacitance is shown as

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Question 182 Marks

In the absence of external electric field, derive the equation of potential energy of a system of three charges.

Answer

$\rightarrow$ Position vectors of three charges $q_1, q_2$ and $q_3$ with respect to the origin are $\vec{r}_1, \vec{r}_2$ and $\vec{r}_3$ respectively.
$\rightarrow$ Suppose these all three charges are situated at infinity in the beginning.
$\rightarrow$ No work is required to bring $q_1$ to $\vec{r}_1$ from an infinite distance.
$( \because$ Electric field is not present.$)$
$\rightarrow$ Work done in bringing charge $q_2$ from infinity to $\vec{r}_2$
$q_2 V_1\left(\vec{r}_2\right)=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r_{12}}$
$\rightarrow$ Charges $q_1$ and $q_2$ both have their own potential at a particular point as well.
Potential at any point $P$ is,
$V _{(1,2)}=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1}{r_{1 p}}+\frac{q_2}{r_{2 p}}\right)$
$\rightarrow$ Work done in bringing charge $q_3$ from infinity to point having position $\vec{r}_3$
Work done $=$ Charge $q_3\ \times $ potential at $q_1$ $q_2$ at position vector $\vec{r}_3$
$=q_3 V_{(1,2)}\left(\overrightarrow{r_3}\right)$
$=q_3\left[\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_{13}}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_{23}}\right]$
$=\frac{1}{4 \pi \varepsilon_0} \cdot\left(\frac{q_1 q_3}{r_{13}}+\frac{q_2 q_3}{r_{23}}\right)$
$\rightarrow$ Equations $(1)$ and $(3)$ are added to get the total work required to be done in bringing all $3$ charges at the given point, which is called the potential energy of the system.
$\therefore U =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r_{12}}+\frac{1}{4 \pi \varepsilon_0} \cdot\left(\frac{q_1 q_3}{r_{13}}+\frac{q_2 q_3}{r_{23}}\right)$
$\therefore U =\frac{1}{4 \pi \varepsilon_0}\left[\frac{q_1 q_2}{r_{12}}+\frac{q_1 q_3}{r_{13}}+\frac{q_2 q_3}{r_{23}}\right] .$
For a system of $n$ charges electric potential energy in the absence of external electric field

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Question 192 Marks

Write down the definition of energy density and derive the equation of energy density of electric field.

Answer

$\rightarrow$ Energy density : "Energy stored per unit volume is called energy density."
$\rightarrow$ Suppose the area of each capacitor plate is $A$ and the distance between plates is $d$.
$\rightarrow$ Energy stored in capacitor.
$U =\frac{ Q ^2}{2 C }$
but $Q =\sigma A \ ($where $\sigma$ is surface charge density$)$
and $C =\frac{\varepsilon_0 A}{d}$
$\therefore U =\frac{\sigma^2 A^2}{2\left(\frac{\varepsilon_0 A}{d}\right)}$
$ =\frac{\sigma^2 Ad }{2 \varepsilon_0} .$
$\rightarrow$ Electric field between two plates,
$E =\frac{\sigma}{\varepsilon_0}$
$ \therefore \sigma= E \varepsilon_0$
From eq. $(1)$ and $(2),$
$\therefore U =\frac{ E ^2 \varepsilon_0^2 Ad }{2 \varepsilon_0}$
$\therefore U =\frac{1}{2} \varepsilon_0 E ^2 A d$
$\therefore \frac{ U }{ A d}=\frac{1}{2} \varepsilon_0 E ^2$
but $A d= V \ ($Volume of the space between two plates$)$
$\therefore \frac{U}{V}=\frac{1}{2} \varepsilon_0 E^2$
$\therefore g_E=\frac{1}{2} \varepsilon_0 E^2$

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Question 202 Marks

Give definition of the smallest unit of energy $, ev\ ($electron volt$)$ and give its explanation.

Answer

$\rightarrow $ Definition : "The change in the energy of an electron which is passed through $($accelerated by$)$ a $p.d. ($potential difference$)$ of $1$ volt is known as one electron volt."
$\rightarrow $ The charge of an electron $q=e=1.6 \times 10^{-19} C$. If it is passed through a potential difference of
$1$ volt, the energy gained by it is,
$=q \Delta V$
$=1.6 \times 10^{-19} \times 1$
$=1.6 \times 10^{-19} J$
Which is called $1 eV . ($electron volt$)$
$\therefore 1 eV =1.6 \times 10^{-19} J$
$\rightarrow$ This unit is generally and most commonly used in atomic and nuclear physics.
$1 keV =1.6 \times 10^{-16} J$
$1 MeV =1.6 \times 10^{-13} J$
$1 GeV =1.6 \times 10^{-10} J$
$1 TeV =1.6 \times 10^{-7} J$

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Question 212 Marks

Explain the process of leakage of electric charge.

Answer

→ Capacitance $C =\frac{ Q }{ V }$
→ The above equation shows that to get a larger value of $C$, for the given value of charge $Q$, value of electric potential should be smaller.
→ This means that a capacitor with a larger capacitance can store large quantity of electric charge for a comparatively smaller electric potential V. But if the p.d. is more, then the electric field in the surrounding region of the conductor is strong.
→ If the air molecules in the surrounding region can be ionized using this strong electric field, the electric charges created by it do motion (/ accelerated motion) towards the capacitor plate, thereby neutralizeing the charge on the capacitor plates.
→ Therefore, the charge of the capacitor leaks away due to the reduction in insulating power of the intervening medium.

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Question 222 Marks

Write the equation of $P.E.$ of electric dipole in external electric field and discuss its special cases.

Answer

$\rightarrow $Electric potential energy of electric dipole in uniform electric field,
$U =-\vec{p} \cdot \overrightarrow{ E }$
$\therefore U =-p E \cos \theta$
$($Where, $\theta$ is the angle between electric dipole moment $\vec{p}$ and electric field $\vec{E}.)$
• Special cases :
$(i)$ Electric dipole is parallel to the electric field
$\therefore \theta=0$
$\therefore U=-p E \cos 0$
$\therefore U=-p E($ minimum $)$
$\rightarrow$ Which shows the stable equilibrium position of dipole.
$(ii)$ If dipole is arranged anti parallel to the electric field, $\theta$ will be equal to $180^{\circ}$. ( $\pi rad$ ).
$\therefore \theta=\pi$
$\therefore U =-p E \cos \pi$
$\therefore U =p E ($ maximum $)\quad(\because \cos \pi=-1)$
$\rightarrow$ Which shows the unstable equilibrium position of dipole.
$(iii)$ If dipole is arranged perpendicular to the electric field then,
$\theta=\frac{\pi}{2}\left(90^{\circ}\right)$
$\therefore U =-p E \cos \frac{\pi}{2}$
$\therefore U =0 \quad\left(\because \cos \frac{\pi}{2}=0\right)$
$\rightarrow$ Which shows the position of maximum nonequilibrium.

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Question 232 Marks

Explain how does the polarized dielectric modify the original external field inside it ?

Answer

→As shown in the figure, consider a rectangular (/ cuboid) dielectric slab placed in a uniform external field $\overrightarrow{ E _0}$.
→Due to this, the dipole moments are aligned almost parallel to $\overrightarrow{ E _0}$.

→The situation shown in the figure is an ideal situation, at 0K (zero kelvin) temperature, The external field causes almost uniform polarization $\overrightarrow{ P }$ of the dielectric, Thus every volume element $\Delta V$ of the slab has a dipole moment $\overrightarrow{ P } \Delta V$.
→Inside the dielectric, the net charge inside the volume element $\Delta V$ is zero. This is because the positive charge of one dipole sits close to and hence cancels out the effect of the negative charge of the neighbouring dipole. However, at the surface of the dielectric net, surface density is not zero because as shown in the figure, the positive ends of the dipoles remain unneutralized at the right surface and the negative ends at the left surface. These unbalanced charges are the induced charges due to the external electric field. Therefore, there is some net surface charge density on the surface of the dielectric.
→The induced electric charge is also known as bound charge.
→Thus the polarized dielectric is equivalent to two charged surfaces with induced surface charge densities, say $\sigma_p$ and $-\sigma_p$.
→The field produced by these surface charges opposes the external electric field. The total field in the dielectric is, there by, reduced.

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Question 242 Marks

Derive the formula for electrostatic potential due to a point charge.

Answer

→As shown in fig., charge $Q(Q>0)$ is placed on the origin of the cartesian co-ordinates system. We want to find electric potential at some point P , having its position vector $\vec{r}$ from origin O. For this we should calculate work done by external force in bringing unit positive charge (test charge) from infinity to point $P$.
→Suppose, there is a point $P ^{\prime}$ at distance $r^{\prime}$ in between the path from infinity to P .
→Force on unit positive charge kept at point $P ^{\prime}$ is
$\overrightarrow{ F }^{\prime}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{ Q }{r^{\prime^2}} \cdot \hat{r}^{\prime}$

Where $\hat{r}^{\prime}$ is the unit vector in the direction of $\overrightarrow{ OP ^{\prime}}$.

→The work done against the (field) force in giving $\overrightarrow{\Delta r^{\prime}}$ displacement to the unit positive charge,
$\Delta W =-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{ Q }{r^{\prime 2}} \cdot \Delta r^{\prime}$

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Question 252 Marks

Write down the equations of electric potential of a point charge and electric potential of dipole. And also discuss (/mention) the points of difference between them.

Answer

→Electric potential at distance ' $r$ ' from point charge,
$V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}$

→Electric potential at a point at distance ' $r$ ' from the mid point of electric dipole,
$V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{p \cos \theta}{r^2}$ Where, $p=$ electric dipole moment
$\theta=$ angle between $\vec{r}$ and $\vec{p}$.

• Points of difference :
(i) Electric potential of dipole depends on distance
$r$ and $\theta$ (which is angle between position vector $\vec{r}$ and dipole moment $\vec{p}$ ), where as electric potential of the point charge depends only on distance $r$.

(ii) Electric potential of dipole reduces with distance as per $\frac{1}{r^2}$, where as electric potential of a point charge reduces with distance as per $\frac{1}{r}$.

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Question 262 Marks

Explain : Electrostatic shielding.

Answer

→Consider a conductor with a cavity as shown in the figure.
→The shape and size of the cavity can be any.
→There is no charge in the cavity.
→ No matter what external field such a conductor is placed in the electric field in the cavity is zero.
→Even if the conductor is charged, all charges, reside only on the outer surface of conductor with cavity.
→Like this, there is no effect of the external electric field or electric charge on the cavity. which means the cavity remains shielded from outside charge and external electric field, which is known as electrostatic shielding.
→This effect can be made of use in protecting sensitive equipments from outside electrical influence.
→E.g. in monsoon season, if we are sitting in a car and lightening strikes, we should close the doors and windows of the car. By doing this, we happen to be in the cavity of car and we are protected due to electro static shielding.

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Question 272 Marks

Prove that the electric field at the surface of a charged conductor is $\overrightarrow{ E }=\frac{\sigma}{\varepsilon_0} \hat{n}$. Where, $\sigma$ - Surface charge density, $\hat{n}$ - unit vector normal to the surface and in going outward direction.

Answer

→To derive the result, choose a pill box (a short cylinder) as the Gaussian surface about any point $P$ on the surface, as shown in the figure.

→The pill-box is partly inside and partly outside the surface of the conductor. It has a small area of cross section $\delta S$ and negligible height.
→ Electro static field just inside the surface is zero. Just outside the surface, field is normal to the surface with magnitude E .
→Thus, the contribution to the total flux through the pill box comes only from outside (circular) cross section of the pill box. Which is equal to $\pm E \delta S$ (Positive for $\sigma>0$, negative for $\sigma<0$.)
→The total charge enclosed by the pill-box is $\delta q=|\sigma| \delta S$

→By Gauss's law,
$\begin{aligned}
E \delta S & =\frac{\delta q}{\varepsilon_0} \\
E \delta S & =\frac{|\sigma| \delta S }{\varepsilon_0} \\
\therefore \quad E & =\frac{|\sigma|}{\varepsilon_0}
\end{aligned}$

→In vector form,
$\overrightarrow{ E }=\frac{\sigma}{\varepsilon_0} \hat{n}$

Where, $\sigma$ - Surface charge density
$\hat{n}$ - Unit vector normal to the surface in the outward direction
→For $\sigma>0$, electric field is normal to the surface in outward direction.
→For $\sigma<0$, electric field is normal to the surface in inward direction.

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Question 282 Marks

Gravitational force or spring force is conservative force - explain.

Answer

→When a body is taken from one point to the other point, using an external force against the force like gravitational force or spring force, the work done on the body is stored in the form of potential energy.
→When this external force is removed, this potential energy is converted into kinetic energy. The sum of kinetic energy and potential energy remains constant in this process. Therefore, these type of forces are called conservative forces.

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Question 292 Marks

Mention important results regarding electrostatics of conductors.

Answer

→Important results regarding electrostatics of conductors are as follows :
(1) Inside a conductor, electric field is zero.
(2) At the surface of a charged conductor, electrostatic field must be normal to the surface at every point.
(3) The interior of a conductor can have no excess charge in the static situation.
(4) Electrostatic potential is constant throughout the volume of the conductor and has the same value inside as on its surface.
(5) Electric field at the surface of a charged conductor
$\overrightarrow{ E }=\frac{\sigma}{\varepsilon_0} \hat{n}$
Where, $\sigma$ is the surface charge density and
$\hat{n}$ is a unit vector normal to the surface in the outward direction.
(6) When a conductor with a cavity is placed in external electric field the cavity in the conductor remains shielded from the outside electric influence. (The field inside cavity is always zero. This is known as electrostatic shielding.)

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Question 302 Marks

Obtain the equation of electric potential energy of a system of two electric charges in external electric field.

Answer

→Suppose two charges $q_1$ and $q_2$ are at infinite distance.
→Work done in bringing $q_1$ from infinity to $\overrightarrow{r_1}$ is $q_1 V\left(\overrightarrow{r_1}\right)$.
→To bring charge $q_2$ from infinity to a point (having position vector $\overrightarrow{r_2}$ ), work needs to be done against external electric field $\overrightarrow{ E }$ as well as against the electric field of $q_1$ charge.
(i) Work done on $q_2$ against the external field
$=q_2 V\left(\vec{r}_2\right)$
(ii) Work done on $q_2$ against the electric field of charge $q_1$
$=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r_{12}}$

Where, $r_{12}$ is the distance between charges $q_1$ and $q_2$ i.e. the work W which needs to be done can be written as follows :
$U =q_1 V\left(\overrightarrow{r_1}\right)+q_2 V\left(\overrightarrow{r_2}\right)+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r_{12}}$

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Question 312 Marks

Obtain the equation of electric energy of a single (solo) charge.

Answer

→According to the definition, electric potential energy, is the work done in bringing a given charge $q$, from infinity to given point against the field force.
→Suppose, electric potential energy of test electric charge is to be found out at point P in the external electric field.
→For this, whatever work needs to be done in bringing the given test charge from infinity to point P against the field force, will be the electric potential energy of $q$ charge at point $P$.
→If electric potential at point P is $V ( P )$, then electric potential energy.
$U ( P )=q V( P ) \quad(\because W = V q=\Delta U )$

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Question 322 Marks

Explain why at the surface of a charged conductor, electric field is normal to the surface at each point.

Answer

→Suppose electric field is not normal to the surface, so then it will have some non-zero component parallel to the surface.
→In these circumstances the free electric charges on the surface experience force and start moving.
→Hence in the stable condition, there should be no tangential component (parallel component) of $\vec{E}$ on the surface of conductor.
→Therefore on the surface of a charged conductor static electric field must be normal to the surface at each point.

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Question 332 Marks

Explain : Static electric field inside a conductor is zero.

Answer

→Consider a conductor, neutral or charged. In metallic conductor, there are positive ions fixed at definite positions and there are free electrons randomly moving in the inter ionic space.
→When such conductor is placed in external electric field $\vec{E}$, these free electrons gather (accumulate) at one end and in the same proportion, positive charge opens up at the other end.
→These charges create electric field inside the conductor, direction of which is opposite to the direction of the external field. When these two electric fields are of equal magnitude, net electric field inside conductor becomes zero.

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Question 342 Marks

From the formula of potential energy of a system of two charges, obtain the equation of $P.E$. of electric dipole in external electric field.

Answer

$\rightarrow$ Potential energy of a system of two charges in external electric field $\vec{E}$ is given by,
$U =q_1 V\left(\overrightarrow{r_1}\right)+q_2 V\left(\overrightarrow{r_2}\right)+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r_{12}}$
$\rightarrow $ For the system of electric dipole,
$q_1 =q, q_2=-q$
$\therefore U =q\left[V\left(\overrightarrow{r_1}\right)- V \left(\overrightarrow{r_2}\right)\right]-\frac{q^2}{4 \pi \varepsilon_0 \times 2 a} \cdot$
here, $r_{12}=2 a$ is the distance between two charges and $\overrightarrow{r_1}$ and $\overrightarrow{r_2}$ are position vectors of charges $+q$ and $-q$ respectively.
$\rightarrow$ Now, the potential difference between position $\overrightarrow{r_1}$ and $\overrightarrow{r_2}$ equals the work done in bringing a unit positive charge against field from $\overrightarrow{r_2}$ to $\overrightarrow{r_1}$, the displacement parallel to the force is $2 a \cos \theta$.
$\therefore V \left(\overrightarrow{r_1}\right)- V \left(\vec{r}_2\right)=- E \times 2 a \cos \theta\ (\because V = E . d )$
From equation $(1),$
$\therefore U =q(- E \times 2 a \cos \theta)-\frac{q^2}{4 \pi \varepsilon_0 2 a}$
$\therefore U =-p E \cos \theta-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{2 a}(\because 2 a q=p)$
$\therefore U =-\vec{p} \cdot \overrightarrow{ E }-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{2 a}$
$\rightarrow$ In the above equation $\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{2 a}$ is constant.
$\rightarrow$ In the equation of potential energy the constant is not significant and it can be neglected.
$\therefore$ Equation $(2)$ can be written as follows :
$U =-\vec{p} \cdot \overrightarrow{ E }$

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Question 352 Marks

$1 F\ ($farad$)$ is a very large unit for practical purposes. Explain with example.

Answer

$\rightarrow$ Suppose,
for a parallel plate capacitor having $1 F$ capacitance, distance between two plates is $1 \ cm$ and area of each plate is $A.$
$\rightarrow$ Capacitance of parallel plate capacitor,
$C =\frac{\varepsilon_0 A}{d}$
$\therefore A =\frac{ C d}{\varepsilon_0}=\frac{1 \times 1 \times 10^{-2}}{8.85 \times 10^{-12}}$
$\therefore A \approx 10^9 m^2$
$\therefore A =10 \times 10^8 m^2$
$\rightarrow$ Considering the shape of plate to be a square,
$\therefore l^2=10 \times 10^8 m^2$
$\therefore l \approx 3 \times 10^4 m$
$\therefore l=30 \ km$
$\rightarrow$ Which is practically impossible.
So $,1 F$ is a unit too large for practice.

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Question 362 Marks

Derive the equation of electric potential due to a system of electric charges.

Answer

$\rightarrow$ A system of $n-$ electric charges is shown in the fig., with reference to the origin, their position vectors are $\overrightarrow{r_1}, \overrightarrow{r_2}, \overrightarrow{r_3} \ldots . . \overrightarrow{r_n}$ and the electric charges are respectively $q_1, q_2, q_3 \ldots . . q_n$.
$\rightarrow$ Electric potential at point $p$ due to charge $q_1$
$V _1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_{1 p}}$
Where, $r_{1 p}$ is the distance between $q_1$ and $p$.
$\rightarrow$ Similarly if potential at point $p$ due to charge $q_2$ is $V _2$ and potential at $p$ due to charge $q_3$ is $V _3$ then,
$V _2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_{2 p}} \text { and } V _3=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_3}{r_{3 p}}$
Where, $r_{2 p}$ is the distance between $q_2$ and $p$.
$r_{3 p}$ is the distance between $q_3$ and $p$.
$\rightarrow$ Similarly we can get potential due to other electric charges
$\rightarrow$ So, the total $($net$)$ potential at point $P$ as per the super position principle,
$ V = V _1+ V _2+ V _3+\ldots . .+ V _n$
$V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_{1 p}}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_{2 p}}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_3}{r_{3 p}}$
$+\ldots . .+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_n}{r_{n p}}$
$V=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1}{r_{1 p}}+\frac{q_2}{r_{2 p}}+\frac{q_3}{r_{3 p}}+\ldots . .+\frac{q_n}{r_{n p}}\right)$
$V =\frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{r_{i p}}$

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Question 372 Marks

Explain how electric conduction becomes possible in metallic conductor and in electrolytic conductors.

Answer

→Conductors possess mobile charge carriers. In metallic conductors, there are free electrons as electric charge carriers.
→In metal electrons in the outer most orbit Valence electrons part away from their atoms and they are free to move. These electrons are free to move within metal, but they are not free to leave metal.
→These free electrons collide with each other and with other ions, and move randomly in different directions.
→The positive ions made of bound electrons and nuclei remain held in their fixed positions.
→In electrolytic conductors, the charge carriers are both positive and negative ions, but the movement of the charge carriers is affected by both the external field as well as the chemical forces.

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Question 382 Marks

Write down the difference between series and parallel connection of capacitors

Answer

Series Connection	Parallel Connection
1. Charge on each capacitor is same.	Charge on each capacitor is different.
2. p.d. across two terminals of each capacitor is different	p.d. across two terminals of each capacitor is same.
3. Equivalent capacitance of series connection $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots$	Equivalent capacitance for parallel connection $C = C _1+ C _2+ C _3+\ldots$
4. Value of equivalent capacitance is smaller than the smallest capacitance	Value of equivalent capacitance is more than the largest (/ greatest) capacitance

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Question 392 Marks

Write the definition of electrostatic potential. Tell whether it is a vector or a scalar quantity ? Also write its SI unit and dimensional formula.

Answer

→"Work done in bringing unit positive charge from infinity to the given point in the electric field against the electric field is called electrostatic potential ( $V$ ) at that point."
OR
→"In the region of static electric field, electric potential at any point means work done by external force in bringing unit positive charge from infinity to that point, without acceleration."
→Electrostatic potential is a scalar quantity.
→Its SI unit is : V (Volt) OR J/C.
→Its DF (Dimensional Formula) is $M ^1 L^2 T^{-3} A^{-1}$

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