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

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.

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} \\)

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.

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\\))

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 } } \\)

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.

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\\)

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

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}\\).

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)\\)

Field Due to Electric Dipole | Formulas, Definition, Examples

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}\\).

example

Electric field of an electric dipole for axial points

Two charges

10 μ C

are placed

5.0 m m

apart.Determine the electric field at a point P on the axis of the dipole

15 c m

away from its centre O on the side of the positive charge.
Solution:
Field at P due to charge

+ 10 μ C

= \frac{1 0 ^{- 5}}{4 π ( 8 . 8 5 4 \times 1 0 ^{- 12} C ^{2} N ^{- 1} m ^{- 2} )} \times \frac{1}{( 1 5 - 0 . 2 5 ) ^{2} \times 1 0 ^{- 4} m ^{2}}

= 4.13 \times 1 0^{6} N C^{- 1}

along BP
Field at P due to charge

- 10 μ C

= \frac{1 0 ^{- 5}}{4 π ( 8 . 8 5 4 \times 1 0 ^{- 12} C ^{2} N ^{- 1} m ^{- 2} )} \times \frac{1}{( 1 5 + 0 . 2 5 ) ^{2} \times 1 0 ^{- 4} m ^{2}}

= 3.86 \times 1 0^{6} N C^{- 1}

along PA
The resultant electric field at P due to the two charges at A and B is

= 2.7 \times 1 0^{5} N C^{- 1}

along BP.
In this example, the ratio OP/OB is quite large (= 60). Thus, we can expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole.For a dipole consisting of charges

q, 2 a

distance apart, the electric field at a distance

r

from the centre on the axis of the dipole has a magnitude

E = \frac{2 p}{4 π ϵ _{0} r ^{3}} (r / a > > 1)

where

p = 2 a q

is the magnitude of the dipole moment.The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from

- q t o q

). Here,

p = 1 0^{- 5} C \times 5 \times 1 0^{- 3} m = 5 \times 1 0^{- 8} C m

E = 2.6 \times 1 0^{5} N C^{- 1}

Along the dipole moment direction AB, which is close to the result obtained earlier.

example

Electric field of an electric dipole for equatorial points

Two charges each of 10 C are placed 5.0 mm apart.Determine the electric field at a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole,

E = \frac{p}{4 π ϵ _{0} r ^{3}} (r / a > > 1)

= \frac{5 \times 1 0 ^{- 8}}{4 π \times ( 8 . 8 5 4 \times 1 0 ^{- 12} C ^{2} N ^{- 1} m ^{- 2} )} \times \frac{1}{( 1 5 ) ^{3} \times 1 0 ^{- 6} m ^{3}}

= 1.33 \times 1 0^{5} N C^{- 1}

The direction of electric field in this case is opposite to the direction of the dipole moment vector.

formula

Electric field due to a dipole

E = \frac{1}{4 π ϵ _{0}} \frac{p}{r ^{3}} 3 c o s^{2} θ + 1

where

θ

is the angle between the distance vector and dipole.

definition

Potential due to a dipole at a general point

Potential due to a dipole is given by:

V \approx \frac{p cos θ}{4 π ε _{o} r ^{2}}

where

p :

electric dipole moment

θ :

angle made by the line joining center of dipole and the point with the dipole moment vector

r :

distance between the center of dipole and the point
Note:
Approximation is made that the length of dipole is negligible as compared to the distance of the point from the dipole.

Field Due to Electric Dipole

example

Electric field of an electric dipole for axial points

example

Electric field of an electric dipole for equatorial points

formula

Electric field due to a dipole

definition

Potential due to a dipole at a general point

REVISE WITH CONCEPTS

Torque on a Dipole Placed in an Electric Field

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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?

example

Electric field of an electric dipole for axial points

example

Electric field of an electric dipole for equatorial points

formula

Electric field due to a dipole

definition

Potential due to a dipole at a general point

REVISE WITH CONCEPTS

Torque on a Dipole Placed in an Electric Field

Quick Summary With Stories

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?