Q: Four charges each equal to \$Q\$ are placed at the four corners of a square and a charge \$q\$ is placed at the centre of the square. If the system is in equilibrium, the value of \$q\$ is :

\$ \textbf {Step 1 : Calculate all the required distances between the charges :} \$Let the side length of the square be \$a\$.Refer Figure, In \$\triangle ACD\$, \$\Rightarrow AD^2 +CD^2 = AC^2\$\$\Rightarrow a^2+ a^2 = AC^2\$ \$\Rightarrow AC = \sqrt{2}a\$However, \$ OC =\dfrac{AC}{2} \$ \$=\cfrac{a}{\sqrt{2}}\$\$ \textbf {Step 2 : Equilibrium of central charge} \$By symmetry, Force on central charge will be equal and opposite due to the diagonally opposite charges, which will cancel each other.Hence, Net force on central charge will always be zero, irrespective of value of charge \$q\$.Therefore, to find value of \$q\$ we have to check equilibrium of any one charge at the corner.\$ \textbf {Step 3 : Force due to all the charges at point C } \$Force on charge at \$C\$ due to \$B\$, \$F_1 = \dfrac{K Q^2}{a^2}\$ Force on charge at \$C\$ due to \$D\$, \$F_2 = \dfrac{K Q^2}{a^2}\$ Force on charge at \$C\$ due to \$A\$, \$F_4 = \dfrac{K Q^2}{AC^2} = \dfrac{KQ^2}{2a^2}\$ Force on charge at \$C\$ due to \$q\$ at centre, \$F_3 = \dfrac{K qQ}{OC^2} = \dfrac{2KqQ}{a^2}\$ \$ \textbf {Step 4 : Apply the equilibrium condition at C :} \$For the system to be in equilibrium, net force acting on charge at C must be zero.So, \$\vec{F_3} + \vec{F_1} + \vec{F_2} + \vec{F_4} =0\$ Resultant of \$F_1\$ and \$F_2\$ (Along OC, by symmetry) = \$\sqrt{F_1^2 +F_2^2} = \sqrt{2} F_1\$ Since \$( |F_1| = |F_2|)\$Also, \$F_3\$ & \$F_4\$ are along OC, Therefore, magnitudes of sum of these forces should be zero. \$\Rightarrow \$ \$|F_3| + \sqrt{2} |F_1| + |F_4| =0\$ \$\Rightarrow \dfrac{2K qQ}{a^2}\$ \$+ \sqrt{2} \dfrac{K Q^2}{a^2}\$ \$+ \dfrac{K Q^2}{2a^2}=0\$ \$\Rightarrow 2q = -(\sqrt{2}Q + \dfrac{Q}{2})\$ \$\Rightarrow q = -\dfrac{Q}{4} (1 + 2\sqrt{2})\$

Q: Derive an expression for the electric field at a point on the equatorial plane of an electric dipole.

Consider an electric dipole AB consisting charges +q and -q separated by distance 2l and pole strength q. Let, P be the point on the equatorial line at distance d from mid-point of the dipole i.e. O as shown in figure 1. Now, the electric field, \$E_1\$ at point P due to north pole along AP is, \$E_1 = \dfrac{1}{4\pi \epsilon_0} \dfrac{q}{AP^2}\$ \$E_1 = \dfrac{1}{4\pi \epsilon_0} \dfrac{q}{(d^2 + l^2)^2}\$ .........(1) (\$\because AP^2 = OA^2 + OP^2\$) Similarly, the electric field, \$E_2\$ at point P due to south pole along PB is, \$E_2 = \dfrac{1}{4\pi \epsilon_0} \dfrac{q}{PB^2}\$\$E_2 = \dfrac{1}{4\pi \epsilon_0} \dfrac{q}{(d^2 + l^2)^2}\$ .........(2) (\$\because PB^2 = OB^2 + OP^2\$)From equations (1) and (2), we get\$E_1 = E_2 = E(say)\$Now, the net electric field \$E_P\$ at P due to the dipole is\$E_P = E_1 + E_2 = 2E\$ ........................(3)Resolving \$E_1\$ and \$E_2\$ into horizontal and vertical components, as shown in figure 2. The vertical components will be canceled and horizontal will be added together. So, we can write \$E_P = 2E \cos \theta \$ .............from(3)\$E_P = 2 \left (\dfrac{1}{4\pi \epsilon_0} \dfrac{q}{(d^2 + l^2)^2}) (\dfrac{l}{\sqrt (d^2 + l^2)}\right )\$ ............(\$\because \cos \theta = \dfrac{l}{\sqrt (d^2 + l^2)}\$)For a short dipole, \$l <<d\$, therefore\$B_P = \dfrac{1}{4\pi \epsilon_0} \dfrac{2ql}{d^3}\$\$B_P = \dfrac{1}{4\pi \epsilon_0} \dfrac{Q}{d^3}\$ ........................(\$\because Q = 2ql\$)

Q: An infinite line charge produces a field of \$9\times 10^{4}\$N/C at a distance of 2 cm. Calculate the linear charge density.

\$\textbf{Step 1: Electric field due to infinite line charge}\$ Electric field due to infinite line charge is given by: \$E = \dfrac{\lambda}{2\pi \in_0 d}\$ \$\Rightarrow \lambda = 2\pi \in_0 E \ d\$ \$....(1)\$Where \$\lambda\$ is the linear charge density and \$d\$ is the perpendicular distance of point from the infinite line charge.\$\textbf{Step 2: Substituting the values }\$ \$\because \dfrac{1}{4\pi \in_0} = 9 \times 10^9 Nm^2C^{-2}\$ \$\Rightarrow 2\pi \in_0 = \dfrac{1}{18\times 10^9 Nm^2C^{-2}}\$From Eq \$(1): \$ \$\lambda = \dfrac{1 \times 9 \times 10^4\ NC^{-1}\times 0.02m}{18\times 10^9\ N m^2 C^{-2} }= 10^{-7}\ Cm^{-1}\$

Q: A charge Q is placed at the centre of the line joining two equal charges q. Show that the system of three charges will be in equilibrium if \$Q=\dfrac{-q}{4}\$.

\$\textbf{Step 1: Equilibrium of central charge Q }\$\$\textbf{[Ref. Fig. 1]}\$For the system to be in equilibrium net force on each charge should be zero.By Symmetry, both outer charges \$q\$ will exert equal and opposite force on \$Q\$, which will cancel out. As shown in Figure 1.Hence Net force on \$Q\$ will always be zero irrespective of value of charge.\$\textbf{Step 2: Equilibrium of outer charge q }\$\$\textbf{[Ref. Fig. 2]}\$Net force on charge \$q\$ at the right, due to other two charges should be zero. \$F_{q} = \dfrac{KQq}{x^2} + \dfrac{Kq^2}{(2x)^2} = 0\$ \$\Rightarrow\ \ \dfrac{KQq}{x^2} = -\dfrac{Kq^2}{4x^2}\$ \$\Rightarrow\ \ Q= \dfrac{-q}{4}\$

Q: A polythene piece rubbed with wool is found to have a negative charge of \$3\times 10^{-7}C\$ .(a) Estimate the number of electrons transferred (from which to which?)(b) Is there a transfer of mass from wool to polythene?

Given,\$q=3\times 10^{-7}C\$\$e=1.6\times 10^{-19}C\$(a) The number of the electron,The charge is quantized, \$q=ne\$\$n=\dfrac{q}{e}=\dfrac{3\times 10^{-7}}{1.6\times 10^{-19}}\$\$n=1.875\times 10^{12}\$ electrons(b) Yes, mass of electron is transferred.We know that, \$m=9.1\times 10^{-31}kg\$The transfer of mass, \$M=nm\$\$M=1.875\times 10^{12}\times 9.1\times 10^{-31}\$\$M=17.0\times 10^{-19}kg\$

Q: Two point charges placed at a certain distance r in air exert a force F on each other. The distance r' at which these charges will exert the same force in medium if dielectric constant k is given by

Two point charge placed at a certain distance \$=r\$force \$=F\$dielectric constant \$=k\$Let, two point charges \${ q }_{ 1 }\$ and \${ q }_{ 2 }\$ are separated \$r\$ distance from each other.Then, force experienced between then,\$F=\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \epsilon }_{ 0 }{ r }^{ 2 } } \$Put, \$F=\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \epsilon }_{ 0 }{ r }^{ 2 } } \$\$\therefore\$ \$\dfrac { 4{ q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \epsilon }_{ 0 }{ r }^{ 2 } } =\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ k\left( 4\pi { \epsilon }_{ 0 }{ r }^{ 2 } \right) } \$\$\Rightarrow \dfrac { 1 }{ { r }^{ 2 } } =\dfrac { 1 }{ 4{ r }^{ 2 }h } \$\$\Rightarrow \dfrac { 1 }{ { r }^{ 2 } } =\dfrac { 1 }{ { kr }^{ 2 } } \$taking square root both sides \$\dfrac { 1 }{ r } -\dfrac { 1 }{ \sqrt { k } { r }^{ 1 } } \$or, \${ r }^{ 1 }=\dfrac { r }{ \sqrt { k } } \$

Q: An electron falls through a distance of \$1.5 cm\$ in a uniform electric field of magnitude \$2 \times {10^4}N/C.\$ Now the direction of the field is reversed keeping the magnitude and unchanged and a proton falls through the same distance. Compute the time of fall in each case, neglecting gravity.

ATQ, \$S=1.5cm=1.5\times { 10 }^{ -2 }m\$ \$u=0m/s\$\${ m }_{ e }\$ : mass of electron \$=9.10\times { 10 }^{ -31 }Kg\$\${ m }_{ p }\$ : mass of proton \$=1.672\times { 10 }^{ -27 }Kg\$from coloumb's force, \$\vec { F } =q\vec { E } \$and given \$\vec { E } =2\times { 10 }^{ 4 }N/c\$force proton \${ \vec { F } }_{ (+) }=1.6\times { 10 }^{ -19 }\times 2\times { 10 }^{ 4 }=3.2\times { 10 }^{ -19 }N\$force on electron, \${ \vec { F } }_{ (-) }=-1.6\times { 10 }^{ -19 }\times 2\times { 10 }^{ 4 }=-3.2\times { 10 }^{ -15 }N\$From Newton's Second law, \$\vec { F } =m\vec { a } \$\$\therefore \$ for proton, \${ a }_{ \left( + \right) }=\dfrac { { \vec { F } }_{ (+) } }{ { m }_{ p } } =1.913\times { 10 }^{ 12 }m/{ s }^{ 2 }\$for electron, \${ a }_{ \left( - \right) }=\dfrac { { \vec { F } }_{ \left( - \right) } }{ { m }_{ e } } =-0.35\times { 10 }^{ 16 }m/{ s }^{ 2 }\$from equ of motion, \$S=ut+\dfrac { 1 }{ 2 } { at }^{ 2 }\quad \longrightarrow (1)\$ATQ, The direction of the field was reversed, hence we will ignore the sign convention for acceleration.\$\therefore \$ Substituting values in equ (1) for proton,\$1.5\times { 10 }^{ -2 }=\dfrac { 1 }{ 2 } \times 1.913\times { 10 }^{ 12 }\times { t }^{ 2 }\$\$1.568\times { 10 }^{ -14 }={ t }^{ 2 }\$\$\Rightarrow \quad t=1.252\times { 10 }^{ -7 }seconds\$Substituting values in equ(1) for electron,\$1.5\times { 10 }^{ -2 }=\dfrac { 1 }{ 2 } \times 3.5\times { 10 }^{ 15 }\times { t }^{ 2 }\$\$0.857\times { 10 }^{ -17 }={ t }^{ 2 }\$\$\Rightarrow \quad { t }_{ 0 }=2.927\times { 10 }^{ -19 }seconds\$

Question 1

Derive an expression for electric field due to an electric dipole at a point on its axial line.

Accepted Answer

Electric field due to an electric dipole at a point on its axial line: $AB$ is an electric dipole of two point charges $-q$ and $+q$ separated by small distance $2d$. $P$ is a point along the axial line of the dipole at a distance $r$ from the midpoint $O$ of the electric dipole.The electric fiedl at the point $P$ due to $+q$ placed at $B$ is,${ E }_{ 1 }=\dfrac { 1 }{ 4\pi { \varepsilon  }_{ 0 } } \dfrac { q }{ { \left( r-d \right)  }^{ 2 } }$ (along $BP$)The electric field at the point $P$ due to $-q$ placed at $A$ is,${ E }_{ 2 }=\dfrac { 1 }{ 4\pi { \varepsilon  }_{ 0 } } \dfrac { q }{ { \left( r+d \right)  }^{ 2 } }$ (along $PA$)Therefore, the magnitude of resultant electric field $\left( E \right)$ acts in the direction of the vector with a greater, magnitude. The resultant electric field at $P$ is$E={ E }_{ 1 }+\left( -{ E }_{ 2 } \right)$$E=\left[ \dfrac { 1 }{ 4\pi { \varepsilon  }_{ 0 } } \dfrac { q }{ { \left( r-d \right)  }^{ 2 } } - \dfrac { 1 }{ 4\pi { \varepsilon  }_{ 0 } } \dfrac { q }{ { \left( r+d \right)  }^{ 2 } }  \right]$ along $BP$$E=\dfrac { q }{ 4\pi { \varepsilon  }_{ 0 } } \left[ \dfrac { 1 }{ { \left( r-d \right)  }^{ 2 } } - \dfrac { 1 }{ { \left( r+d \right)  }^{ 2 } }  \right]$ along $BP$$E=\dfrac { q }{ 4\pi { \varepsilon  }_{ 0 } } \left[ \dfrac { 4rd }{ { \left( { r }^{ 2 }-{ d }^{ 2 } \right)  }^{ 2 } }  \right]$ along $BP$If the point $P$ is far away from the dipole, then $d \ll r$$\therefore E=\dfrac { q }{ 4\pi { \varepsilon  }_{ 0 } } - \dfrac { 4rd }{ { r }^{ 4 } } = \dfrac { q }{ 4\pi { \varepsilon  }_{ 0 } } \dfrac { 4d }{ { r }^{ 3 } }$$E=\dfrac { 1 }{ 4\pi { \varepsilon  }_{ 0 } } \dfrac { 2p }{ { r }^{ 3 } }$ along $BP$             [$\because $ Electric dipole moment $p=q\times 2d$]$E$ acts in the direction of dipole moment.

Question 2

Derive an expression for electric potential at a point due to an electric dipole. Discuss the special cases.

Accepted Answer

Let an electric dipole consist of two equal and opposite point charges $&#8211; q$ at A and $+q$ at B ,separated by a small distance AB =$2a$ ,with centre at O.The dipole moment, $p=q\times2a$We will calculate potential at any point P, where$OP=r$ and  $\angle BOP= \theta$Let $BP=r_1$ and $AP=r_2$Draw AC perpendicular PQ and BD perpendicular POIn $\Delta AOC$ $cos\theta=OC/OA =OC/a$$OC= a cos \theta$Similarly,  $  OD= a cos \theta$Potential at P due to $+q=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r_2}$And  Potential at P due to $-q=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r_1}$Net potential at P due to the dipole$V=\dfrac{1}{4\pi\epsilon_0}(\dfrac{q}{r_2}-\dfrac{q}{r_1})$$\implies V=\dfrac{q}{4\pi\epsilon_0}(\dfrac{1}{r_2}-\dfrac{1}{r_1})$Now, $r_1=AP=CP$$=OP+OC$$=r+a cos\theta$And $r_2=BP=DP$$=OP &#8211;OD$$=r-acos\theta$$V=\dfrac{q}{4\pi\epsilon_0}(\dfrac{1}{r-acos\theta}-\dfrac{1}{r+acos\theta})$$=\dfrac{q}{4\pi\epsilon_0}(\dfrac{2acos\theta}{r^2-a^2cos^2\theta})$$=\dfrac{pcos\theta}{r^2-a^2cos^2\theta}$(Since $p=2aq$)Special cases:-(i) When the point P lies on the axial line of the dipole, $\theta=0^{\circ}$$cos\theta=1$$V=\dfrac{p}{r^2-a^2}$If  $a<<r$, $V=\dfrac{p}{r^2}$Thus due to an electric dipole ,potential, $V\propto \dfrac{1}{r^2}$(ii) When the point P lies on the equatorial line of the dipole, $\theta=90^{\circ}$$cos\theta=0$i.e electric potential due to an electric dipole is zero at every point on the equatorial line of the dipole.

Question 3

Four charges each equal to $Q$ are placed at the four corners of a square and a charge $q$ is placed at the centre of the square. If the system is in equilibrium, the value of $q$ is :

Accepted Answer

$ \textbf {Step 1  : Calculate all the required distances between the charges :} $Let the side length of the square be $a$.Refer Figure, In $\triangle ACD$, $\Rightarrow AD^2 +CD^2 = AC^2$$\Rightarrow a^2+ a^2 = AC^2$                 $\Rightarrow  AC = \sqrt{2}a$However, $ OC =\dfrac{AC}{2} $ $=\cfrac{a}{\sqrt{2}}$$ \textbf {Step 2 : Equilibrium of central charge} $By symmetry, Force on central charge will be equal and opposite due to the diagonally opposite charges, which will cancel each other.Hence, Net force on central charge will always be zero, irrespective of value of charge $q$.Therefore, to find value of $q$ we have to check equilibrium of any one charge at the corner.$ \textbf {Step 3 : Force due to all the charges at point C } $Force on charge at $C$ due to $B$,        $F_1 = \dfrac{K Q^2}{a^2}$ Force on charge at $C$ due to $D$,        $F_2 = \dfrac{K Q^2}{a^2}$   Force on charge at $C$ due to  $A$,        $F_4 = \dfrac{K Q^2}{AC^2}  = \dfrac{KQ^2}{2a^2}$       Force on charge at $C$ due to  $q$ at centre,     $F_3 = \dfrac{K qQ}{OC^2}  = \dfrac{2KqQ}{a^2}$       $ \textbf {Step 4 : Apply the equilibrium condition at C :} $For the system to be in equilibrium, net force acting on charge at C must be zero.So,   $\vec{F_3} + \vec{F_1} + \vec{F_2}  + \vec{F_4} =0$     Resultant of $F_1$ and $F_2$ (Along OC, by symmetry) = $\sqrt{F_1^2 +F_2^2} = \sqrt{2} F_1$            Since $( |F_1| = |F_2|)$Also, $F_3$ & $F_4$ are along OC, Therefore, magnitudes of sum of these forces should be zero. $\Rightarrow  $           $|F_3| + \sqrt{2}  |F_1| + |F_4| =0$ $\Rightarrow  \dfrac{2K qQ}{a^2}$   $+ \sqrt{2} \dfrac{K Q^2}{a^2}$ $+ \dfrac{K Q^2}{2a^2}=0$     $\Rightarrow 2q =  -(\sqrt{2}Q + \dfrac{Q}{2})$              $\Rightarrow  q = -\dfrac{Q}{4}  (1 + 2\sqrt{2})$

Question 4

Derive an expression for the electric field at a point on the equatorial plane of an electric dipole.

Accepted Answer

Consider an electric dipole AB consisting charges +q and -q separated by distance 2l and pole&#10;strength q. Let, P&#10;be the point on the equatorial line at distance d from mid-point of the dipole i.e.&#10;O as shown in figure 1.&#10;&#10;Now,&#10; the electric field, $E_1$ at point P due to north pole along AP is,&#10;&#10;$E_1&#10;= \dfrac{1}{4\pi \epsilon_0} \dfrac{q}{AP^2}$&#10;&#10;$E_1 = \dfrac{1}{4\pi \epsilon_0} \dfrac{q}{(d^2 + l^2)^2}$&#10;.........(1)         ($\because AP^2 = OA^2 + OP^2$)&#10;&#10;Similarly,  the electric field, $E_2$ at point P due to&#10;south pole along PB is,&#10;&#10;$E_2 = \dfrac{1}{4\pi \epsilon_0} \dfrac{q}{PB^2}$$E_2 = \dfrac{1}{4\pi \epsilon_0} \dfrac{q}{(d^2 + l^2)^2}$ .........(2)            ($\because PB^2 = OB^2 + OP^2$)From equations (1) and (2), we get$E_1 = E_2 = E(say)$Now, the net electric field $E_P$ at P due to the dipole is$E_P = E_1 + E_2 = 2E$ ........................(3)Resolving $E_1$ and $E_2$ into horizontal and vertical components, as shown in figure 2. The vertical components will be canceled and horizontal will be added together. So, we can write $E_P = 2E \cos \theta $  .............from(3)$E_P = 2 \left (\dfrac{1}{4\pi \epsilon_0} \dfrac{q}{(d^2 + l^2)^2}) (\dfrac{l}{\sqrt (d^2 + l^2)}\right )$ ............($\because \cos \theta = \dfrac{l}{\sqrt (d^2 + l^2)}$)For a short dipole, $l <<d$, therefore$B_P = \dfrac{1}{4\pi \epsilon_0} \dfrac{2ql}{d^3}$$B_P = \dfrac{1}{4\pi \epsilon_0} \dfrac{Q}{d^3}$ ........................($\because Q = 2ql$)

Question 5

An infinite line charge produces a field of $9\times 10^{4}$N/C at a distance of 2 cm. Calculate the linear charge density.

Accepted Answer

$\textbf{Step 1: Electric field due to infinite line charge}$ Electric field due to infinite line charge is given by:   $E = \dfrac{\lambda}{2\pi \in_0 d}$    $\Rightarrow \lambda = 2\pi \in_0 E \ d$                                  $....(1)$Where $\lambda$ is the linear charge density and $d$ is the perpendicular distance of point from the infinite line charge.$\textbf{Step 2: Substituting  the values }$                     $\because \dfrac{1}{4\pi \in_0} = 9 \times 10^9 Nm^2C^{-2}$  $\Rightarrow 2\pi \in_0 = \dfrac{1}{18\times 10^9 Nm^2C^{-2}}$From Eq $(1): $     $\lambda = \dfrac{1 \times 9 \times 10^4\ NC^{-1}\times 0.02m}{18\times 10^9\ N m^2 C^{-2} }= 10^{-7}\ Cm^{-1}$

Question 6

A charge Q is placed at the centre of the line joining two equal charges q. Show that the system of three charges will be in equilibrium if $Q=\dfrac{-q}{4}$.

Accepted Answer

$\textbf{Step 1: Equilibrium of central charge Q }$$\textbf{[Ref. Fig. 1]}$For the system to be in equilibrium net force on each charge should be zero.By Symmetry, both outer charges $q$ will exert equal and opposite force on $Q$, which will cancel out. As shown in Figure 1.Hence Net force on $Q$ will always be zero irrespective of value of charge.$\textbf{Step 2: Equilibrium of outer charge q }$$\textbf{[Ref. Fig. 2]}$Net force on charge $q$ at the right, due to other two charges should be zero.               $F_{q} = \dfrac{KQq}{x^2} + \dfrac{Kq^2}{(2x)^2} = 0$           $\Rightarrow\ \ \dfrac{KQq}{x^2} = -\dfrac{Kq^2}{4x^2}$           $\Rightarrow\ \ Q= \dfrac{-q}{4}$

Question 7

A polythene piece rubbed with wool is found to have a negative charge of $3\times 10^{-7}C$ .(a) Estimate the number of electrons transferred (from which to which?)(b) Is there a transfer of mass from wool to polythene?

Accepted Answer

Given,$q=3\times 10^{-7}C$$e=1.6\times 10^{-19}C$(a) The number of the electron,The charge is quantized, $q=ne$$n=\dfrac{q}{e}=\dfrac{3\times 10^{-7}}{1.6\times 10^{-19}}$$n=1.875\times 10^{12}$ electrons(b) Yes, mass of electron is transferred.We know that, $m=9.1\times 10^{-31}kg$The transfer of mass, $M=nm$$M=1.875\times 10^{12}\times 9.1\times 10^{-31}$$M=17.0\times 10^{-19}kg$

Question 8

Two point charges placed at a certain distance r in air exert a force F on each other. The distance r' at which these charges will exert the same force in medium if dielectric constant k is given by

Accepted Answer

Two point charge placed at a certain distance $=r$force $=F$dielectric constant $=k$Let, two point charges ${ q }_{ 1 }$ and ${ q }_{ 2 }$ are separated $r$ distance from each other.Then, force experienced between then,$F=\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \epsilon  }_{ 0 }{ r }^{ 2 } } $Put, $F=\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \epsilon  }_{ 0 }{ r }^{ 2 } } $$\therefore$   $\dfrac { 4{ q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \epsilon  }_{ 0 }{ r }^{ 2 } } =\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ k\left( 4\pi { \epsilon  }_{ 0 }{ r }^{ 2 } \right)  } $$\Rightarrow \dfrac { 1 }{ { r }^{ 2 } } =\dfrac { 1 }{ 4{ r }^{ 2 }h } $$\Rightarrow \dfrac { 1 }{ { r }^{ 2 } } =\dfrac { 1 }{ { kr }^{ 2 } } $taking square root both sides $\dfrac { 1 }{ r } -\dfrac { 1 }{ \sqrt { k } { r }^{ 1 } } $or,  ${ r }^{ 1 }=\dfrac { r }{ \sqrt { k }  } $

Question 9

An electron falls through a distance of $1.5 cm$ in a uniform electric field of magnitude $2 \times {10^4}N/C.$ Now the direction of the field is reversed keeping the magnitude and unchanged and a proton falls through the same distance. Compute the time of fall in each case, neglecting gravity.

Accepted Answer

ATQ,  $S=1.5cm=1.5\times { 10 }^{ -2 }m$              $u=0m/s$${ m }_{ e }$ : mass of electron $=9.10\times { 10 }^{ -31 }Kg$${ m }_{ p }$ : mass of proton $=1.672\times { 10 }^{ -27 }Kg$from coloumb's force, $\vec { F } =q\vec { E } $and given $\vec { E } =2\times { 10 }^{ 4 }N/c$force proton ${ \vec { F }  }_{ (+) }=1.6\times { 10 }^{ -19 }\times 2\times { 10 }^{ 4 }=3.2\times { 10 }^{ -19 }N$force on electron, ${ \vec { F }  }_{ (-) }=-1.6\times { 10 }^{ -19 }\times 2\times { 10 }^{ 4 }=-3.2\times { 10 }^{ -15 }N$From Newton's Second law,  $\vec { F } =m\vec { a } $$\therefore $  for proton, ${ a }_{ \left( + \right)  }=\dfrac { { \vec { F }  }_{ (+) } }{ { m }_{ p } } =1.913\times { 10 }^{ 12 }m/{ s }^{ 2 }$for electron, ${ a }_{ \left( - \right)  }=\dfrac { { \vec { F }  }_{ \left( - \right)  } }{ { m }_{ e } } =-0.35\times { 10 }^{ 16 }m/{ s }^{ 2 }$from equ of motion, $S=ut+\dfrac { 1 }{ 2 } { at }^{ 2 }\quad \longrightarrow (1)$ATQ, The direction of the field was reversed, hence we will ignore the sign convention for acceleration.$\therefore $  Substituting values in equ (1) for proton,$1.5\times { 10 }^{ -2 }=\dfrac { 1 }{ 2 } \times 1.913\times { 10 }^{ 12 }\times { t }^{ 2 }$$1.568\times { 10 }^{ -14 }={ t }^{ 2 }$$\Rightarrow \quad t=1.252\times { 10 }^{ -7 }seconds$Substituting values in equ(1) for electron,$1.5\times { 10 }^{ -2 }=\dfrac { 1 }{ 2 } \times 3.5\times { 10 }^{ 15 }\times { t }^{ 2 }$$0.857\times { 10 }^{ -17 }={ t }^{ 2 }$$\Rightarrow \quad { t }_{ 0 }=2.927\times { 10 }^{ -19 }seconds$

Question 10

State Gauss' law. Using this find an expression for electric field due to an infinitely long straight charged wire uniform charge density.

Accepted Answer

Gauss's law :The law relates the flux through any closed surface and the net charge enclosed with in the surface. The law states that the total flux of the electric field $E$ over any closed surface is equal to $\dfrac { 1 }{ { \varepsilon  }_{ 0 } }$ times the net charge enclosed by the surface.$\phi =\dfrac { q }{ { \varepsilon  }_{ 0 } }$This closed imaginary surface is called Gaussian surface. Gauss's law tells us that the flux of $E$ through a closed surface $S$ depends only on the value of net charge inside the surface and not on the location of the charges. Charges outside the surface will not contribute to flux.Electric field due to an infinitely long straight uniformly charged wire :Consider an uniformly charged wire of infinite length having a constant linear charge density $\lambda$ (Charge per unit length). Let $P$ be a point at a distance $r$ from the wire and $E$ be the electric field at the point $P$. A cylinder of length $l$, radius $r$, closed at each end by plane caps normal to the axis is chosen as Gaussian surface. Consider a very small area $ds$ on the Gaussian surface. By symmetry, the magnitude of the electric field will be the same at all points on the curved surface of the cylinder and directed radially outward. $\vec { E }$ and $\vec { ds }$ are along the same direction.The electric flux $\left( \phi  \right)$ through curved surface $=\displaystyle\oint { Eds\cos { \theta  }  }$$\phi =\displaystyle\oint { Eds }$             $ \left[ \because \theta =0;\cos { \theta  } =1 \right]$  $=E\left( 2r\pi l \right)$ [The surface area of the curved part is ] since $\vec { E }$ and $\vec { ds }$ are right angles $2\pi rl$ to each other, the electric flux through the plane caps $=0$.$\therefore$ Total flux through the Gaussian surface, $\phi =E\left( 2\pi rl \right)$. The net charge enclosed by Gaussian surface is, $q=\lambda l$$\therefore$ By Gauss's law,$ =E\left( 2\pi rl \right) \dfrac { \lambda l }{ { \varepsilon  }_{ 0 } }$ or $ E=\dfrac { \lambda  }{ 2\pi { \varepsilon  }_{ 0 }r }$The direction of electric field $E$ is radially outward, if line charge is positive and inward, if the line charge is negative.