Video Lecture
Theory For Making Notes
Electric dipole
Two equal and opposite charges separated by a finite distance constitute an electric dipole. The figure shows a dipole with two charges -q and +q separated by a distance 2a.
The dipole moment \vec{p} is given as \vec{p}=q\times 2\vec{a} with a direction from negative to positive charge and \left| p \right|=2qa
It has dimension [LTA].
The SI unit of dipole moment is Cm (coulomb x meter) and experimental unit is Debye.
Electric field due to a dipole at axial point
Let the charges -q and +q are kept at point (-a, 0) & (a, 0) respectively in xy plane. The electric field at point P(x, 0) will be then the vector sum of the two electric fields produced by the two charges of the dipole
\displaystyle {{\vec{E}}_{{axial}}}= \displaystyle {{\vec{E}}_{{+q}}}+{{\vec{E}}_{{-q}}}
or \displaystyle {{\vec{E}}_{axial}}= \frac{{Kq}}{{{{{\left( {x-a} \right)}}^{{_{2}}}}}}\hat{i}-\frac{{Kq}}{{{{{\left( {x+a} \right)}}^{2}}}}\hat{i} , where \displaystyle \hat{i} is the unit vector along axis.
\displaystyle {{\vec{E}}_{{axial}}}=Kq.\frac{{\left( {{{{\left( {x+a} \right)}}^{2}}-{{{\left( {x-a} \right)}}^{2}}} \right)}}{{{{{\left( {{{x}^{2}}-{{a}^{2}}} \right)}}^{2}}}}\hat{i}
or \displaystyle {{\vec{E}}_{{axial}}}=\frac{{2K(q2a)x}}{{{{{({{x}^{2}}-{{a}^{2}})}}^{2}}}}\hat i
since p=q.2a hence finally we get
\displaystyle {{\vec{E}}_{{axial}}}=\frac{{2Kp.x}}{{{{{\left( {{{x}^{2}}-{{a}^{2}}} \right)}}^{2}}}}\hat{i}
Special Case
For a short dipole x>>a hence neglecting a from the denominator we get \displaystyle \,{{\vec{E}}_{{axial}}}=\frac{{2K\vec{p}}}{{{{x}^{3}}}}
Note that the net electric field I parallel to the dipole moment
Electric field on equatorial line or perpendicular bisector of dipole
At P there are two electric fields due to two charges of dipole.
\displaystyle {{\vec{E}}_{{+q}}}=\frac{{Kq}}{{\left( {{{y}^{2}}+{{a}^{2}}} \right)}}(-\cos \theta \hat{i}+\sin \theta \hat{j})
\displaystyle {{\vec{E}}_{{-q}}}=\frac{{Kq}}{{\left( {{{y}^{2}}+{{a}^{2}}} \right)}}(-\cos \theta \hat{i}-\sin \theta \hat{j});
The net electric field \displaystyle \,\,{{\vec{E}}_{P}} is the sum of the two vectors hence
\displaystyle \,\,{{\vec{E}}_{P}}={{\vec{E}}_{{+q}}}+{{\vec{E}}_{{-q}}}=\frac{{-2Kq}}{{\left( {{{y}^{2}}+{{a}^{2}}} \right)}}\cos \theta \hat{i}
\displaystyle \Rightarrow {{\vec{E}}_{p}}=-2\left[ {\frac{{Kq}}{{{{y}^{2}}+{{a}^{2}}}}} \right].\frac{a}{{{{{\left( {{{a}^{2}}+{{y}^{2}}} \right)}}^{{{}^{1}\!\!\diagup\!\!{}_{2}\;}}}}}\hat{i}
\displaystyle \Rightarrow {{\vec{E}}_{p}} =\frac{{-K\left( {2aq} \right)}}{{{{{\left( {{{y}^{2}}+{{a}^{2}}} \right)}}^{{{}^{3}\!\!\diagup\!\!{}_{2}\;}}}}}\hat{i}
\displaystyle\Rightarrow {{\vec{E}}_{p}}= \frac{{-K\vec{p}}}{{{{{({{y}^{2}}+{{a}^{2}})}}^{{3/2}}}}}
The negative sign indicates that the direction of net electric field is opposite to that of dipole moment `p`.
Electric field at any point due to a short dipole
Consider a dipole of moment \vec{p} and a point P at distance r where field is required. Now, the \vec{p} has two components p\cos \theta and p\sin \theta .
Note that the point P is an axial point for p\cos \theta component and is an equatorial point for p\sin \theta component.
Therefore, the electric field at P due to p\cos \theta component is
{{E}_{1}}=\frac{{2kp\cos \theta }}{{{{r}^{3}}}} …(i)
and the electric field at P due to p\sin \theta component is
{{E}_{2}}=\frac{{kp\sin \theta }}{{{{r}^{3}}}} …(ii)
Since \displaystyle {{E}_{1}}{ } and{ } {{E}_{2}} are mutually perpendicular so the net electric field is given by
\displaystyle {{E}_{{net}}}=\sqrt{{E_{1}^{2}+E_{2}^{2}}}
\displaystyle \Rightarrow{ }{{E}_{{net}}}=\sqrt{{{{{\left( {\frac{{2kp\cos \theta }}{{{{r}^{3}}}}} \right)}}^{2}}+{{{\left( {\frac{{kp\sin \theta }}{{{{r}^{3}}}}} \right)}}^{2}}}}
\Rightarrow{ }{{E}_{net}}=\frac{{kp}}{{{{r}^{3}}}}\sqrt{{4{{{\cos }}^{2}}\theta +{{{\sin }}^{2}}\theta }}
Hence {{E}_{net}}=\frac{{kp}}{{{{r}^{3}}}}\sqrt{{3{{{\cos }}^{2}}\theta +1}}
\tan \alpha =\frac{{{{E}_{2}}}}{{{{E}_{1}}}}=\frac{1}{2}\tan \theta
Dipole in electric field
When a dipole is placed in a uniform field as shown in figure. The net force on it is zero.
Force, F=qE is acting on each charge but in opposite direction so they form a couple.
Net torque = moment of the couple
or Torque = Force x arm of couple
therefore \tau=qE\times 2a\sin \theta
Hence \tau =pE\sin \theta
\because \,\,\,\,\,\vec{\tau }=\vec{p}\times \vec{E}
Note
(i)
when \theta =0, \tau =0 ,it is called the stable equilibrium position of dipole
(ii)
when \theta =90{}^\circ , torque is maximum and given by {{\tau }_{{\max }}}=PE
(iii)
when \theta =180{}^\circ , \tau =0 , it is called unstable equilibrium position of dipole
Work done in rotating a dipole in uniform field
Let a dipole is placed in an electric field at an initial angle θ1. It is to be rotated to a new position where it makes an angle θ2 with the electric field. Now let us rotate the dipole through a small angle d\theta . If dW be the work done, it will be given as dW=\tau \,\,d\,\,\theta
to get the total work done we integrate it as follows
\int{{dW}}=\int_{{{{\theta }_{1}}}}^{{{{\theta }_{2}}}}{{\,pE\sin \theta \,\,d\theta }}
W=-pE[\cos {{\theta }_{1}}-\cos {{\theta }_{2}}]
Since this work is done against the electric field by some external agent so it will change the potential energy of the system.
Hence W=\Delta U where \Delta U is the change in potential energy
Therefore \Delta U =-pE[\cos {{\theta }_{1}}-\cos {{\theta }_{2}}]
Special cases for clculating work
using the equation W=-pE[\cos {{\theta }_{1}}-\cos {{\theta }_{2}}]
1.
If the dipole moment were initially parallel to electric field and then rotated to a position making an angle \theta with the electric field.
hence {{\theta }_{1}}=0 and {{\theta }_{2}}=\theta ,
then W=pE[1-\cos \theta ]
2.
If the dipole moment were initially parallel to electric field and then rotated to a position making an angle {{180}^{o}} with the electric field.
hence {{\theta }_{1}}=0 and {{\theta }_{2}}={{180}^{o}}
then W=pE[cos(0) – cos {{180}^{o}} ] this is the maximum work required to rotate the dipole.
therefore {{W}_{{\max }}}=2PE
Potential Energy: In case of a dipole, potential energy is defined as work done in rotating the dipole from a direction perpendicular to the field to the given final position. Actually the potential energy of a dipole is considered to be zero when it is perpendicular to eletric field.
we know that \Delta U =-pE[\cos {{\theta }_{1}}-\cos {{\theta }_{2}}]
where \Delta U = {{U}_{final}} – {{U}_{initial}}
hence {{U}_{final}} – {{U}_{initial}}=-pE[\cos {{\theta }_{1}}-\cos {{\theta }_{2}}]
now {{\theta }_{1}}=90{}^\circ ,\,\,{{\theta }_{2}}=\theta
so {{U}_{final}} – {{U}_{initial}}=pE(cos90{}^\circ -\cos \theta ) here {{U}_{initial}}=0
So finally we get {{U}_{final}}=-pE\cos \theta
or {{U}_{final}}=-\vec{p}\,.\vec{E}
So the potential energy of a dipole in an external field is given by
\displaystyle U=-\vec{p}\,.\,\vec{E}
The potential energy as a function of the angle \theta is shown in figure.
The minimum potential energy occurs at \theta={{0}^{\circ}},and is given by U=-pE
Whereas the maximum potential energy comes at \theta=\pi and is given by U=pE.
At \theta={{0}^{\circ}} the dipole is said to be in stable equilibrium position.
If the dipole from the stable equilibrium position is rotated slightly and then left free , it starts oscillating about the direction of the electric field.
Force acting on a dipole in an External Non-uniform field.
The charges of a dipole experience unequal forces, therefore, the net force on the dipole is not equal to zero. The magnitude of the force is given by the negative derivative of the potential energy with respect to distance along the axis of the dipole.
F = \displaystyle -\frac{{dU}}{{dx}}
F = – \displaystyle p.\frac{{dE}}{{dx}}
Illustration
A dipole is formed by two charges of 5μC and -5μC at a distance of 8 mm. Find electric field at
(a) a point `A’ 25 cm away from dipole centre at its axis.
(b) a point `B’ 20 cm away on perpendicular to the axis and passing through its centre
Solution
(a)
E = \displaystyle \frac{{2p}}{{4\pi {{\varepsilon }_{0}}{{r}^{3}}}} = \displaystyle \frac{{2\times 5\times {{{10}}^{{-6}}}\times 8\times {{{10}}^{{-3}}}\times 9\times {{{10}}^{9}}}}{{{{{\left( {25} \right)}}^{3}}\times {{{10}}^{{-6}}}}} = 4.6 x 104 N/C
(b)
E = – \displaystyle \frac{p}{{4\pi {{\varepsilon }_{0}}{{r}^{3}}}} = \displaystyle -\frac{{5\times {{{10}}^{{-6}}}\times 8\times {{{10}}^{{-3}}}\times 9\times {{{10}}^{9}}}}{{{{{\left( {20} \right)}}^{3}}\times {{{10}}^{{-6}}}}} = – 4.5 x 104 N/C
Illustration
Two tiny spheres, each of mass M, and charges +q and -q respectively, are connected by a massless rod of length, L. They are placed in a uniform electric field at an angle \theta with the \vec{E} . Calculate the minimum time in which the system aligns itself parallel to the \vec{E}.(Assuming \displaystyle \theta \to 0).
We know that
\tau=pEsin\theta ,
as \displaystyle \theta \to 0, sin \theta \to \theta
therefore \tau=-pE\theta
here minus sign signifies that If we assume angular displacement to be anti-clockwise, torque is clockwise means the angular diplacement and torque are in opposite direction.
Also we know I\alpha=\tau , where I is the moment of inertia and \alpha is the angular acceleration
hence I\alpha=-pE\theta
or \displaystyle \alpha= -\left( {\frac{{pE}}{I}} \right)\theta this relation of \alpha shows that the motion of the system is S.H.M.
we also know that the standard equation of \alpha in S.H.M. is given by
\displaystyle \alpha=-{{\omega}^{2}}.\theta
when we compare the two equations of angular acceleration we get
\displaystyle-{{\omega}^{2}}.\theta = -\left( {\frac{{pE}}{I}} \right)\theta
or \displaystyle \omega =\sqrt{{\frac{{pE}}{I}}}
or \displaystyle \frac{{2\pi }}{T}=\sqrt{{\frac{{pE}}{I}}}
therefore time period T is given by
\displaystyle T=2\pi \sqrt{{\frac{I}{{pE}}}}
Here, p = qL and moment of inertia I is given by
\displaystyle I=M{{\left( {\frac{L}{2}} \right)}^{2}}+M{{\left( {\frac{L}{2}} \right)}^{2}}=\frac{{M{{L}^{2}}}}{2}
Hence the time to rotate from angle \theta to a position parallel to \vec E is t=\frac{T}{4}
so \displaystyle t=\frac{\pi }{2}\sqrt{{\frac{I}{{pE}}}}
Force acting between two short dipoles when they are placed coaxially
Two short dipoles are placed along the same line as shown in the figure. We can say that the dipole {{p}_{2}} is placed in the nonuniform field of the dipole {{p}_{1}}.
The electric field of {{p}_{1}} on its axial line is \displaystyle E=\frac{{2K{{p}_{1}}}}{{{{r}^{3}}}}.
differentiating E w.r.t. r we get
\displaystyle \frac{{dE}}{{dr}}=-\frac{{6K{{p}_{1}}}}{{{{r}^{4}}}}
Now the force on the dipole {{p}_{2}} in the nonuniform electric field of {{p}_{1}} is given by
\displaystyle F=-{{p}_{2}}\frac{{d{{E}_{1}}}}{{dr}}
Hence \displaystyle F=\frac{{6K{{p}_{2}}{{p}_{1}}}}{{{{r}^{4}}}}
Potential Energy Of the system of two dipoles placed coaxially
Since \displaystyle F=-\frac{{dU}}{{dr}}
so \displaystyle dU=-F.dr,
on integrating we get \displaystyle \int{{dU=-\int{{F.dr}}}}
so \displaystyle U=-\int{{\frac{{6K{{p}_{2}}{{p}_{1}}}}{{{{r}^{4}}}}.dr}},
FInally \displaystyle U=\frac{{2K{{p}_{2}}{{p}_{1}}}}{{{{r}^{3}}}}
Illustration
Find the net dipole moment of a system of three charges arranged as shown in figure.
Solution
As shown in the figure given below the system is suppose to have two dipoles of dipole moment p each. Where p=q.l
The vector sum of these two dipole moments is the resultant dipole moment of the system. To find the resultant each dipole moment is resolved into its two rectangular components i.e. p.sin\theta and p.cos\theta as shown in the figure. (Here \theta={{30}^{\circ}}). It can be easily seen that p.sin\theta components will cancel each other
whereas p.cos\theta components will be added.
So resultant dipole moment {{p}_{net}} is given as
{{p}_{net}}= 2p.cos\theta
finally {{p}_{net}}= 2p.cos{{30}^{\circ}}
or \displaystyle {{p}_{{net}}}=2p\frac{{\sqrt{3}}}{2}=\sqrt{3}(ql)
Note : The same resultant can be calculated by using law of parallelogram.
Hence
\displaystyle {{p}_{{net}}}=\sqrt{{{{p}^{2}}+{{p}^{2}}+2p.p\cos 60}}=\sqrt{3}p=\sqrt{3}(ql)
Practice Questions (Basic Level)
Q. 1
An electric dipole placed in a uniform electric field experiences, in general
(a) a force and a torque
(b) a force only
(c) a torque only
(d) neither a force nor a torque
Ans. (c)
Q.2
A and B are two points on the axis and the perpendicular bisector respectively of an electric dipole. And B are far away from the dipole and at equal distances from it. The fields at A and B are{{E}_{A}}{ }and{ }{{E}_{B}}.
(a) {{E}_{A}}{ }={ }{{E}_{B}}
(b) {{E}_{A}}{ }={ }{2{E}_{B}}
(c) {{E}_{A}}{ }={ }{-2{E}_{B}}
(d) {{E}_{B}}{ }={ }{-2{E}_{A}}
Ans. (c)
Q.3
An electric dipole of moment \vec p is placed in a uniform electric field \vec E, with \vec p parallel to \vec E . It is then rotated by an angle \theta. The work done is
(a) \vec p \times \vec E
(b) \vec p . \vec E
(c) pE- \vec p . \vec E
(d) pE- \vec p\times \vec E
Ans. (c)
Q. 4
In which of the following states is the potential energy of an electric dipole maximum ?
Ans (a)
Q.5
An electric dipole of moment p is placed normal to the lines of force of electric field E, then the work done in deflecting it through an angle of 180 degrees is
(a) pE
(b) + 2pE
(c) – 2pE
(d) Zero
Ans. (d)
Q.6
A dipole is oscillating in an electric field with a time period of 1sec. If its moment of inertia is \frac{1}{{{\pi}^{2}}}{ }kg{{m}^{2}}. Then the maximum kinetic energy of the dipole during oscillation must be
(a) 2J
(b) 4J
(c) zero
(d) 6J
Ans: (b)
Q.7
An electric dipole is put in north-south direction in a sphere filled with water. Which statement is correct?
(a) Electric flux is coming towards sphere
(b) Electric flux is coming out of sphere
(c) Electric flux entering into sphere and leaving the sphere are same
(d) Water does not permit electric flux to enter into sphere.
Ans. (c)
Q.8
A dipole is present in a uniform electric field with the dipole parallel to field. Now the dipole is rotated by {{30}^{\circ}} and released. The subsequent motion of dipole is
(a) simple harmonic but not periodic
(b) periodic but not simple harmonic
(c) neither simple harmonic nor periodic
(d) simple harmonic and periodic.
Ans. (b)
Q.9
For a dipole \displaystyle q=2\times {{10}^{{-6}}}C and d = 0.01 m. Calculate the maximum torque for this dipole if \displaystyle E=5\times {{10}^{5}}N/C
(a) \displaystyle 1\times {{10}^{{-3}}}N{{m}^{{-1}}}
(b) \displaystyle 10\times {{10}^{{-3}}}N{{m}^{{-1}}}
(c) \displaystyle 10\times {{10}^{{-3}}}Nm
(d) \displaystyle 1\times {{10}^{2}}N{{m}^{2}}
Ans. (c)
Q.10
The force between any two charged bodies
(a) is always proportional to square of the distance between them
(b) is always inversly proportional to square of the distance between them
(c) is independent of the distance between them
(d) depends on the distance raise to power n, where the value of n depends on the charged bodies.
Ans : (d)
Practice Questions (JEE Main Level)
Q.1
A short electric dipole is placed at the orgin O and is directed along the x-axis. At a point P, far away from the dipole, the electric field is parallel to the y-axis. If OP makes an angle q with the x-axis,then
(a) tan q=Ö3 (b) tan q= Ö2 (c) q= 45° (d) tan q=1/Ö2
Ans. (b)
Q.2
Electric charges q, q and -2q are placed at the corners of an equilateral triangle ABC of side L. The magnitude of electric dipole moment of the system is
(a) qL (b) 2qL (c) \sqrt{3}.qL (d) 4qL
Ans. (c)
Q.3
An electric dipole is kept in non-uniform electric field. It may experience
(a) A force and a torque
(b) A force but not a torque
(c) A torque but not a force
(d) option (a) and (b) both
Ans. (d)
Q.4
Three electric charges 2q each are placed at the corners of an equilateral triangle ABC of side l. Another charge -6q is placed at the centroid of the triangle. The magnitude of electric dipole moment of the system is
(a) q
(b) \displaystyle 6ql \frac {1}{\sqrt{3}}
(c) \displaystyle \sqrt{3}ql
(d) zero
Ans. (d)
Q.5
If the magnitude of intensity of electric field at a distance x on axial line and at a distance y on equatorial line on a given dipole are equal, then x : y is
(a) 1 : 1
(b) \displaystyle 1:\sqrt{2}
(c) 1 : 2
(d) \displaystyle \sqrt[3]{2}:1
Ans. (d)
Q.6
An electric dipole is placed along the x-axis at the origin O. A point P is at a distance of 20cm from this origin such that OP makes an angle \displaystyle \frac{\pi }{3} with the x-axis. If the electric filed at P makes an angle θ with x-axis, the value of θ would be
(a) \displaystyle \frac{\pi }{3}
(b) \displaystyle \frac{\pi }{3}+{{\tan }^{{-1}}}\left( {\frac{{\sqrt{3}}}{2}} \right)
(c) \displaystyle \frac{{2\pi }}{3}
(d) \displaystyle {{\tan }^{{-1}}}\left( {\frac{{\sqrt{3}}}{2}} \right)
Ans. (b)
Q.7
Two short electric dipoles of moment P and 64 P are placed in opposite direction on a line at a distance of 25 cm. The electric field will be zero at point between the dipoles whose distance from the dipole of moment P is
(a) 5 cm
(b) \displaystyle \frac{{25}}{9}cm
(c) 10 cm
(d) \displaystyle \frac{4}{{13}}cm
Ans. (a)
Q.8
Two short dipoles \displaystyle p\,\hat{k} and \displaystyle \frac{p}{2}\,\hat{k} are located at (0, 0, 0) & (1m, 0, 2m) respectively. The resultant electric field due to the two dipoles at the point (1m, 0, 0) is
(a) \displaystyle \frac{{9p}}{{32\,\pi \,{{\varepsilon }_{0}}}}\,\hat{k}
(b) \displaystyle \frac{{-7p}}{{32\,\pi \,{{\varepsilon }_{0}}}}\,\hat{k}
(c) \displaystyle \frac{{7p}}{{32\,\pi \,{{\varepsilon }_{0}}}}\,\hat{k}
(d) None of these
Ans. (b)
Q.9
An electric dipole is kept on the axis of a uniformly charged ring at distance \frac{R}{\sqrt2} from the centre of the ring. The direction of the dipole moment is along the axis. The dipole moment is p, charge of the ring is Q & radius of the ring is R. The force on the dipole is
(a) \displaystyle \frac{{p\,Q}}{{3\,\pi \,{{\varepsilon }_{0}}\sqrt{3}\,{{R}^{2}}}}
(b) \displaystyle \frac{{4p\,Q}}{{3\,\pi \,{{\varepsilon }_{0}}\sqrt{3}\,{{R}^{2}}}}
(c) \displaystyle \frac{{p\,Q}}{{3\,\pi \,{{\varepsilon }_{0}}\,{{R}^{2}}}}
(d) zero
Ans. (d)
Q.10
The force acting between two dipoles is proportional to xn . Where x is their mutual distance and n is
(a) -4 (b) -3 (c) 4 (d) -2
Ans. (a)
Q.11
A solid nonconducting sphere has a uniform charge density \rho. A spherical cavity is made in the solid sphere whose position vector of the center makes {{60}^{\circ}} with the x axis as shown in the diagram. A short dipole is placed in this cavity in stable equilibrium. Which of the following may be the dipole moment of this dipole
(a) \displaystyle \vec{p}=\hat{i}+2\hat{j}
(b) \displaystyle \vec{p}=\sqrt{2}\hat{i}+\sqrt{2}\hat{j}
(c) \displaystyle \vec{p}=\sqrt{2}\hat{i}+\sqrt{6}\hat{j}
(d) \displaystyle \vec{p}=3\hat{i}+\sqrt{2}\hat{j}
Ans: (c)
Q.12
If the electric field is uniform in all cases. Choose the correct statement for these dipoles
(a) All are in rotatory equilibrium
(b) In all cases the potential energy is non zero
(c)All are in translatory as well as rotatory equilibrium
(d) All are only in translatory equilibrium.
Ans. (d)
Q.13
Two connected charges of +q and -q respectively are at fixed distance AB apart in a non uniform electric field whose lines of force are shown in the figure
The resultant effect on the two charges is
(a) a torque vector in the plane of the paper and no resultant force.
(b) a resultant force in the plane of the paper and no torque.
(c) a torque vector normal to the plane of the paper and no resultant force.
(d) a torque vector normal to the plane of the paper and a resultant force in the plane of the paper.
Ans. (d)
Q.14
We have two electric dipoles. Each dipole consists of two equal and opposite point charges at the end of an insulating rod of length d.
The dipoles are placed along the x‑axis at a large distance r apart oriented as shown. The dipole on the left
(a) will feel a force upwards and a torque trying to make it rotate clockwise.
(b) will feel a force upwards and a torque trying to make it rotate counterclockwise.
(c) will feel a force upwards and no torque about its centre.
(d) will feel a force downwards and a torque trying to make it rotate clockwise.
Ans. (b)
Q.15
Two short dipoles with moments \vec {{p}_{1}}, \vec {{p}_{2}} are lying in x-z plane with their centers coinciding with each other at a point having coordinates (2,0,4). An electron {{e}^{-}} is placed at (2,d,4). If \vec {{p}_{1}}=-p\hat k and \vec {{p}_{2}}=p\hat i then the force vector (\vec F) on the electron must be
(a) \displaystyle \vec{F}=\frac{{Kpe}}{{{{d}^{2}}}}(\hat{i}+\hat{k})
(b) \displaystyle \vec{F}=\frac{{Kpe}}{{{{d}^{3}}}}(-\hat{i}-\hat{k})
(c) \displaystyle \vec{F}=\frac{{2Kpe}}{{{{d}^{3}}}}(\hat{i}+\hat{j})
(d) \displaystyle \vec{F}=\frac{{Kpe}}{{{{d}^{3}}}}(\hat{i}-\hat{k})
Ans. (d)
16.
Half part of ring is uniformly positively charged and other half is uniformly negatively charged. Ring is in equilibrium in uniform electric field as shown and free to rotate about an axis passing through its centre and perpendicular to plane. The equilibrium is
(a) In the given situation the ring can be in unstable equilibrium
(b) The ring is in equilibrium with minimum potential energy and zero torque
(c) The ring is in equilibrium with maximum potential energy
(d) If the ring is given a small angular displacement in its own plane and released it will remain at rest at new position
Ans (b)
17.
The magnitude of electric field intensity at point B (2, 0, 0) due to a dipole of dipole moment, \vec{P}=\hat{i}+\sqrt{3}\hat{j} kept at origin is (assume that the point B is at large distance from the dipole and k=\frac{1}{{4\pi {{\varepsilon }_{0}}}})
(a) \displaystyle \frac{{\sqrt{{13}}k}}{8}
(b) \frac{{\sqrt{{13}}k}}{4}
(c) \frac{{\sqrt{7}k}}{8}
(d) \frac{{\sqrt{7}k}}{4}
Ans (c)
18.
A dipole of dipole moment p is kept at the centre of a ring of radius R and charge density \lambda. The dipole moment has direction along the axis of the ring. The resultant force on the ring due to the dipole is
(a) zero
(b) \frac{{p\lambda \pi }}{{2\pi {{\varepsilon }_{0}}{{R}^{2}}}}
(c) \frac{{p\lambda \pi}}{{4\pi {{\varepsilon }_{0}}{{R}^{2}}}}
(d) \frac{{p\lambda \pi}}{{4\pi {{\varepsilon }_{0}}{{R}^{3}}}}only if the charge is uniformly distributed on the ring.
Ans (b)
Comprehension Based Questions (19 and 20)
A dipole consists of two point charges ±2 nC separated by 4 cm.
19.
What is the dipole moment ?
(a) 9 x 10–11 C-m
(b) 8 x 10–11 C-m
(c) 6 x 10–11 C-m
(d) 3 x 10–11 C-m
Ans (b)
20.
What is the change in potential energy when the dipole rotates from alignment along a field E={{10}^{5}}{ }N/C to an orientation {{90}^{\circ}} to E?
(a) 2 x 10–6 J
(b) 8 x 10–6 J
(c) 10 x 10–6 J
(d) 5 x 10–6 J
Ans (b)
Practice Questions (JEE Advance Level)
Q.1
Imagine a dipole is at the centre of a spherical surface. If magnitude of electric field at a certain point on the surface of sphere is 10 N/C, then which of the following cannot be the magnitude of electric field anywhere on the surface of sphere
(a) 4 N/C (b) 8 N/C (c) 16 N/C (d) 18 N/C
Ans. (a)
Q.2
Two positive point charges are kept along x-axis on each side of origin in x-y plane at An electric dipole of length a is kept along y-axis with its mid point at origin such that its dipole moment is along +ve y-axis. The dipole is released from given position then
(a) the dipole will vibrate in SHM along y-axis with mean position at origin
(b) the dipole will move linearly along +y axis
(c) the dipole will be in rotational as well as translation motion
(d) the dipole will oscillate on y-axis
Ans. (b)
Q.3
A non-conducting ring of mass m and radius r is lying at rest in the vertical xy plane on a smooth non-conducting horizontal xz plane. Charges +q and –q are distributed uniformly on the two sides of the vertical diameter of the ring, and a constant and uniform electric field \displaystyle \vec{E} is set up along the x-direction. The ring is given a small rotation about the vertical diameter of the ring and released. Find the period of oscillation of the ring.
(a) \displaystyle 2\pi \sqrt{{\frac{{mr\pi}}{{8qE}}}}
(b) \displaystyle 2\pi \sqrt{{\frac{{mr}}{{10qE}}}}
(c) \displaystyle \pi \sqrt{{\frac{{3mr}}{{4qE}}}}
(d) \displaystyle 2\pi \sqrt{{\frac{{mr}}{{16qE}}}}
Ans (a)
4.
An electric dipole is placed along the x-axis at the origin O. A point P is at a distance of 20cm from the origin such that OP makes an angle \frac{\pi }{3} with the x-axis. If the electric field at P makes an angle \theta with the positive direction of x-axis, the value of \theta would be
(a) \frac{\pi }{3}
(b) \frac{\pi }{3}+{{\tan }^{{-1}}}\left( {\frac{{\sqrt{3}}}{2}} \right)
(c) \frac{{2\pi }}{3}
(d) {{\tan }^{{-1}}}\left( {\frac{{\sqrt{3}}}{2}} \right)
Ans (b)
Comprehension Based Question (5 and 6)
The figure shows a combination of charges called an electric quadrupole. Find the electric field strength
5.
At point A at (r, 0)
(a) \frac{{2kq}}{{{{r}^{2}}}} \left[ {5-{{{\left( {2+\frac{{{{a}^{2}}}}{{{{r}^{2}}}}} \right)}}^{{-3/2}}}} \right]
(b) \frac{{2kq}}{{{{r}^{2}}}} \left[ {3-{{{\left( {5+\frac{{{{a}^{2}}}}{{{{r}^{2}}}}} \right)}}^{{-3/2}}}} \right]
(c) \frac{{2kq}}{{{{r}^{2}}}} \left[ {9-{{{\left( {3+\frac{{{{a}^{2}}}}{{{{r}^{2}}}}} \right)}}^{{-3/2}}}} \right]
(d) \frac{{2kq}}{{{{r}^{2}}}} \left[ {1-{{{\left( {1+\frac{{{{a}^{2}}}}{{{{r}^{2}}}}} \right)}}^{{-3/2}}}} \right]
Ans (d)
6.
At point B at (0, r)
(a) \frac{{5kq{{a}^{3}}({{a}^{2}}-3{{r}^{2}})}}{{{{r}^{2}}{{{({{r}^{2}}-{{a}^{2}})}}^{2}}}}
(b) \frac{{9kq{{a}^{2}}({{a}^{5}}-3{{r}^{2}})}}{{{{r}^{2}}{{{({{r}^{2}}-{{a}^{5}})}}^{2}}}}
(c) \frac{{2kq{{a}^{2}}({{a}^{2}}-3{{r}^{2}})}}{{{{r}^{2}}{{{({{r}^{2}}-{{a}^{2}})}}^{2}}}}
(d) \frac{{2kq{{a}^{2}}({{a}^{2}}-3{{r}^{2}})}}{{{{r}^{2}}{{{({{r}^{4}}-{{a}^{4}})}}^{4}}}}
Ans (c)