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 $\dfrac{1}{4\pi&#10;{{\varepsilon }_{0}}}\left( \dfrac{2P}{\left( {{r}^{2}}-4{{a}^{2}} \right)}&#10;\right)$

Introduction to Electric Dipole and Electric Dipole Moment

Language

Rate

REVISE WITH CONCEPTS

Electric Dipole

Field Due to Electric Dipole

QUICK SUMMARY WITH STORIES

Electric Dipole and Dipole Moment

Electric Field Due to a Dipole

Electric Field Due to a Dipole (Axial Point)

Electric Field Due to a Dipole (Equatorial Point)

Electric Field Due to Dipole at Any Point - Polar Coordinate

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

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

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

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

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

An electric dipole with dipole moment $4 \times 1 0^{- 9}$ C m is aligned at $3 0^{\circ}$ with the direction of a uniform electric field of magnitude $5 \times 1 0^{4} N C^{- 1}$ . Calculate the magnitude of the torque acting on the dipole.

An electric dipole consist of two opposite charges each of magnitude $1 μ C$ separated by a distance of $2 c m .$ The dipole is placed in an external field of $1 0^{5} N/C$ .The maximum torque on the dipole is:

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 :

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

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

Introduction to Electric Dipole and Electric Dipole Moment

Language

Rate

REVISE WITH CONCEPTS

Electric Dipole

Field Due to Electric Dipole

QUICK SUMMARY WITH STORIES

Electric Dipole and Dipole Moment

Electric Field Due to a Dipole

Electric Field Due to a Dipole (Axial Point)

Electric Field Due to a Dipole (Equatorial Point)

Electric Field Due to Dipole at Any Point - Polar Coordinate

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

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

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

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

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

An electric dipole with dipole moment 4×10−9C m is aligned at 30∘ with the direction of a uniform electric field of magnitude 5×104NC−1. Calculate the magnitude of the torque acting on the dipole.

An electric dipole consist of two opposite charges each of magnitude 1μC separated by a distance of 2cm. The dipole is placed in an external field of 105N/C.The maximum torque on the dipole is:

An electric dipole is placed at an angle of 30o with an electric field intensity 2×105 N/C. It experiences a torque equal to 4 Nm. The charge on the dipole, if the dipole length is 2 cm, is :

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

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

An electric dipole with dipole moment $4 \times 1 0^{- 9}$ C m is aligned at $3 0^{\circ}$ with the direction of a uniform electric field of magnitude $5 \times 1 0^{4} N C^{- 1}$ . Calculate the magnitude of the torque acting on the dipole.

An electric dipole consist of two opposite charges each of magnitude $1 μ C$ separated by a distance of $2 c m .$ The dipole is placed in an external field of $1 0^{5} N/C$ .The maximum torque on the dipole is:

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 :

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