Q: Derive an expression for electric field due electric dipole at a point on an equatorial line.

Electric field at A due to the charges \$q\$ and \$-q\$ is shown in the above figure. We resolve \$E\$ into horizontal and vertical components. The vertical components \$(E \sin\theta)\$ cancel out each other and only the horizontal components survive to give net electric field at A as \$2E\cos\theta\$.Electric field at point A, \$E_A = 2E \cos\theta\$where \$E = \dfrac{q}{4\pi \epsilon_o (r^2 + a^2)}\$We get \$E_A = \dfrac{2q }{4\pi \epsilon_o (r^2 + a^2)} \cos\theta\$From figure, we have \$\cos\theta = \dfrac{a}{\sqrt{r^2 + a^2}}\$\$\therefore\$ \$E_A = \dfrac{2q a}{4\pi \epsilon_o (r^2 + a^2)^{3/2}} \$We know dipole moment \$P = 2qa\$\$\implies\$ \$E_A = \dfrac{P}{4\pi \epsilon_o (r^2 + a^2)^{3/2}} \$For \$a<<r\$, we can neglect \$a^2 \$ compared to \$r^2\$.\$\implies\$ \$E_A = \dfrac{P}{4\pi \epsilon_o r^3} \$

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

Let an electric dipole consist of two equal and opposite point charges \$– 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 –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.

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: 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 electric dipole with dipole moment \$4\times 10^{-9}\$C m is aligned at \$30^\circ\$ with the direction of a uniform electric field of magnitude \$5\times 10^{4} NC^{-1}\$. Calculate the magnitude of the torque acting on the dipole.

Electric dipole moment, \$ E = 4 \times 10^{-9} N/C\$Angle made by p with a uniform electric field, θ = 30°Electric field, \$ E = 5 \times 10^4 N/C\$Torque acting on the dipole: \$\tau=pE\sin \theta\$\$=4\times 10^{-9}\times 5\times 10^4\times sin\, 30^o\$\$=20\times 10^{-5}\times \frac{1}{2}\$\$=10^{-4}Nm\$Hence, the magnitude of the torque acting on the dipole is \$10^{-4}\$ N m.

Q: An electric dipole consist of two opposite charges each of magnitude \$1\mu C\$ separated by a distance of \$2\,cm.\$ The dipole is placed in an external field of \${10^5}{\text{N/C}}\$.The maximum torque on the dipole is:

An electric dipole consist at two opposite charge each of magnitude \$=1\mu C=1\times { 10 }^{ -6 }C\$distance \$=2cm\$Exter field \$={ 10 }^{ 5 }N/C\$maximum torque on the dipole \$=?\$\$q=1\times { 10 }^{ -6 }C,\quad 2a=2cm\$ or, \$=0.02cm\$\$\therefore\$ \$P=q\times 2a\$ \$=\left( 1\times { 10 }^{ -6 } \right) \times 0.02\$ \$=2\times { 10 }^{ -8 }cm\$Intensity of the external electric field, \$E=1.0\times { 10 }^{ 5 }N/C\$(i) \${ Z }_{ max }=pE=\left( 2\times { 10 }^{ -8 } \right) \left( 10\times { 10 }^{ 5 } \right) =2\times { 10 }^{ -3 }N-m\$(ii) Net work done in turning the dipole from \${ 0 }^{ 0 }\$ to \${ 180 }^{ 0 }\$i.e \$W=\int _{ { 0 }^{ 0 } }^{ { 180 }^{ 0 } }{ \overline { r } } d\theta =\int _{ { 0 }^{ 0 } }^{ { 180 }^{ 0 } }{ pE\sin\theta } d\theta \$ \$=pE{ \left[ -cos\theta \right] }_{ { 0 }^{ 0 } }^{ { 180 }^{ 0 } }\$ \$=-pE\left( { \cos180 }^{ 0 }-\cos{ 0 }^{ 0 } \right) \$ \$=2pE\$ \$=2\times \left( 2\times { 10 }^{ -8 } \right) \left( 1\times { 10 }^{ 5 } \right) J\$ \$=4\times { 10 }^{ -3 }J\$

Q: An electric dipole is placed at an angle of \$30^o\$ with an electric field intensity \$2\times 10^5\ N/C\$. It experiences a torque equal to \$4\ Nm.\$ The charge on the dipole, if the dipole length is \$2\ cm\$, is :

\$\textbf{Step 1: Dipole moment of dipole}\$Given the dipole length \$l = 2\ cm = 2/100\ m\$So, the Dipole moment of dipole will be \$P=lq\$ \$ =(2/100)q\$ \$P=0.02q\$\$\textbf{Step 2: Torque acting on dipole }\$ \$\vec \tau = \vec P \times \vec E \$\$\Rightarrow \ \ \tau=PE\sin\theta\$ Or, \$4\ Nm=(0.02q)(2\times 10^5\ N/C)\sin30^o\$Or, \$q=200 \times 10^{-5}C\$ \$ =2 mC\$Hence, Option C is correct

Q: Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.

Consider the an electric dipole of charges \$+q\$ and \$-q\$ separated by distance \$2a\$ with center at O. Goal: To find electric field at point P on the axial line of the dipole, \$OP=r\$. Let \$E_1\$ and \$E_2\$ be electric field on P due to charges \$+q\$ and \$-q\$ respectively. \$E_1 = \dfrac{kq}{(r-a)^2}\$ along AP \$E_2 = \dfrac{kq}{(r+a)^2}\$ along BPThe resultant electric field at P, \$E=E_1 - E_2\$ (as both \$E_1\$ and \$E_2\$ are in opposite direction) \$E=\dfrac{kq}{(r-a)^2}-\dfrac{kq}{(r+a)^2} \\ =kq \dfrac{4ra}{(r^2 - a^2)^2}\$Define, \$p=2aq\$ \$E=k \dfrac{2pr}{(r^2 - a^2)^2}\$If \$r>>a\$, then \$E=k \dfrac{2pr}{r^4} = k\dfrac{2p}{r^3}\$. In vector form, \$\overrightarrow{E}=k\dfrac{2\overrightarrow{p}}{r^3}\$.

Q: Derive the expression for the electric potential at anypoint along the axial line of an electric dipole?

Let P be an axial point at distance r from the center of the dipole. Electric potential at point P is given as \$V={{V}_{1}}+{{V}_{2}}\$ \${{V}_{1}}\$and \${{V}_{2}}\$are the potentials at point P due to charges +q and –q respectively \$ V=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{q}{r-2a}+\left( \dfrac{-q}{r+2a} \right) \right) \$ \$ V=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{4a}{\left( {{r}^{2}}-4{{a}^{2}} \right)} \right) \$ \$ V=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{2P}{\left( {{r}^{2}}-4{{a}^{2}} \right)} \right) \$ Hence, the electric potential is \$\dfrac{1}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{2P}{\left( {{r}^{2}}-4{{a}^{2}} \right)} \right)\$

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 field due electric dipole at a point on an equatorial line.

Accepted Answer

Electric field at A due to the charges $q$ and $-q$ is shown in the above figure. We resolve $E$ into horizontal and vertical components. The vertical components $(E \sin\theta)$ cancel out each other and only the horizontal components survive to give net electric field at A as $2E\cos\theta$.Electric field at point A, $E_A = 2E \cos\theta$where $E = \dfrac{q}{4\pi \epsilon_o (r^2 + a^2)}$We get $E_A = \dfrac{2q }{4\pi \epsilon_o (r^2 + a^2)} \cos\theta$From figure, we have $\cos\theta = \dfrac{a}{\sqrt{r^2 + a^2}}$$\therefore$ $E_A = \dfrac{2q a}{4\pi \epsilon_o (r^2 + a^2)^{3/2}} $We know dipole moment $P = 2qa$$\implies$ $E_A = \dfrac{P}{4\pi \epsilon_o (r^2 + a^2)^{3/2}} $For $a<<r$, we can neglect $a^2 $ compared to $r^2$.$\implies$ $E_A = \dfrac{P}{4\pi \epsilon_o r^3} $

Question 3

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 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

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 6

An electric dipole with dipole moment $4\times 10^{-9}$C m is aligned at $30^\circ$ with the direction of a uniform electric field of magnitude $5\times 10^{4} NC^{-1}$. Calculate the magnitude of the torque acting on the dipole.

Accepted Answer

Electric dipole moment, $ E = 4 \times 10^{-9} N/C$Angle made by p with a uniform electric field, &#952; = 30&#176;Electric field, $ E = 5 \times 10^4 N/C$Torque acting on the dipole: $\tau=pE\sin \theta$$=4\times 10^{-9}\times 5\times 10^4\times sin\, 30^o$$=20\times 10^{-5}\times \frac{1}{2}$$=10^{-4}Nm$Hence, the magnitude of the torque acting on the dipole is $10^{-4}$ N m.

Question 7

An electric dipole consist of two opposite charges each of magnitude $1\mu C$ separated by a distance of $2\,cm.$ The dipole is placed in an external field of ${10^5}{\text{N/C}}$.The maximum torque on the dipole is:

Accepted Answer

An electric dipole consist at two opposite charge each of magnitude $=1\mu C=1\times { 10 }^{ -6 }C$distance $=2cm$Exter field $={ 10 }^{ 5 }N/C$maximum torque on the dipole $=?$$q=1\times { 10 }^{ -6 }C,\quad 2a=2cm$                                or,  $=0.02cm$$\therefore$    $P=q\times 2a$           $=\left( 1\times { 10 }^{ -6 } \right) \times 0.02$           $=2\times { 10 }^{ -8 }cm$Intensity of the external electric field, $E=1.0\times { 10 }^{ 5 }N/C$(i) ${ Z }_{ max }=pE=\left( 2\times { 10 }^{ -8 } \right) \left( 10\times { 10 }^{ 5 } \right) =2\times { 10 }^{ -3 }N-m$(ii) Net work done in turning the dipole from ${ 0 }^{ 0 }$ to ${ 180 }^{ 0 }$i.e  $W=\int _{ { 0 }^{ 0 } }^{ { 180 }^{ 0 } }{ \overline { r }  } d\theta =\int _{ { 0 }^{ 0 } }^{ { 180 }^{ 0 } }{ pE\sin\theta  } d\theta $           $=pE{ \left[ -cos\theta  \right]  }_{ { 0 }^{ 0 } }^{ { 180 }^{ 0 } }$           $=-pE\left( { \cos180 }^{ 0 }-\cos{ 0 }^{ 0 } \right) $           $=2pE$           $=2\times \left( 2\times { 10 }^{ -8 } \right) \left( 1\times { 10 }^{ 5 } \right) J$           $=4\times { 10 }^{ -3 }J$

Question 8

An electric dipole is placed at an angle of $3 0^{o}$ with an electric field intensity $2 \times 1 0^{5} N / C$ . It experiences a torque equal to $4 N m .$ The charge on the dipole, if the dipole length is $2 c m$ , is :

Accepted Answer

$\textbf{Step 1: Dipole moment of dipole}$Given the dipole length $l = 2\ cm = 2/100\ m$So, the Dipole moment of dipole will be           $P=lq$  $ =(2/100)q$           $P=0.02q$$\textbf{Step 2: Torque acting on dipole }$ $\vec \tau = \vec P \times \vec E $$\Rightarrow \ \ \tau=PE\sin\theta$ Or,   $4\ Nm=(0.02q)(2\times 10^5\ N/C)\sin30^o$Or,   $q=200 \times 10^{-5}C$ $ =2 mC$Hence, Option C is correct

Question 9

Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.

Accepted Answer

Consider the an electric dipole of charges $+q$ and $-q$ separated by distance $2a$ with center at O. Goal: To find electric field at point P on the axial line of the dipole, $OP=r$. Let $E_1$ and $E_2$ be electric field on P due to charges $+q$ and $-q$ respectively. $E_1 = \dfrac{kq}{(r-a)^2}$ along AP $E_2 = \dfrac{kq}{(r+a)^2}$ along BPThe resultant electric field at P, $E=E_1 - E_2$ (as both $E_1$ and $E_2$ are in opposite direction) $E=\dfrac{kq}{(r-a)^2}-\dfrac{kq}{(r+a)^2} \ =kq \dfrac{4ra}{(r^2 - a^2)^2}$Define,  $p=2aq$ $E=k \dfrac{2pr}{(r^2 - a^2)^2}$If $r>>a$, then $E=k \dfrac{2pr}{r^4} = k\dfrac{2p}{r^3}$. In vector form, $\overrightarrow{E}=k\dfrac{2\overrightarrow{p}}{r^3}$.

Question 10

Derive the expression for the electric potential at anypoint along the axial line of an electric dipole?

Accepted Answer

Let P be an axial point at distance r from&#10;the center of the dipole. Electric potential at point P is given as&#10;&#10;$V={{V}_{1}}+{{V}_{2}}$ &#10;&#10;${{V}_{1}}$and ${{V}_{2}}$are the&#10;potentials at point P due to charges +q and &#8211;q respectively&#10;&#10;  $ V=\dfrac{1}{4\pi&#10;{{\varepsilon }_{0}}}\left( \dfrac{q}{r-2a}+\left( \dfrac{-q}{r+2a} \right)&#10;\right) $ &#10;&#10; $ V=\dfrac{1}{4\pi&#10;{{\varepsilon }_{0}}}\left( \dfrac{4a}{\left( {{r}^{2}}-4{{a}^{2}} \right)}&#10;\right) $ &#10;&#10; $ V=\dfrac{1}{4\pi&#10;{{\varepsilon }_{0}}}\left( \dfrac{2P}{\left( {{r}^{2}}-4{{a}^{2}} \right)}&#10;\right) $ &#10;&#10;Hence, the electric potential is $\dfrac{1}{4\pi&#10;{{\varepsilon }_{0}}}\left( \dfrac{2P}{\left( {{r}^{2}}-4{{a}^{2}} \right)}&#10;\right)$

Electric Field Due to a Dipole (Axial Point)

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Field Due to Electric Dipole

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