Q: Derive the formula for the electric potential energy of an electric dipole in a uniform electric field.

Potential energy of dipole in uniform electric field.\$\vec {\tau}=\vec p\times \vec E\$work done in rotating dipole by small angle\$dw=\tau d\theta\$\$=-p\varepsilon \sin \theta d\theta\$The change in potential energy \$dU=-dW=pE\sin \theta d\theta\$If \$\theta \rightarrow 90^o\$ to \$\theta^o\$\$U(\theta) -U(90^o)=\displaystyle \int_{90^o}^{0^o}{P\varepsilon \sin \theta d\theta}\$\$=P\varepsilon [-\cos \theta]_{90}^{0}\$\$=-\vec P.\vec E\$So\$\boxed{U(\theta)=-\vec P. \vec E}\$

Q: An electric dipole of length 4cm when placed with its axis making an angle of \${60^ \circ }\$ with a uniform electric field, experience a torque of \$4\sqrt 3 N - m\$. Calculate the potential energy of the dipole of it has charge \$ \pm 8nC\$.

Dipole moment(P) = q.2L\$P=8\times { 10 }^{ -9 }\times 2\times 4\times { 10 }^{ -2 }\\ \quad =64\times { 10 }^{ -11 }Cm\$Torque on the dipole in uniform electric field\$\tau =PEsin\theta \\ E=\dfrac { \tau }{ Psin\theta } \$Potential energy of the dipole is \$U=-PEcos\theta \$on substituting the value , we get the value of E and U\$U=-P\dfrac { \tau }{ Psin\theta } cos\theta \\ \quad \quad =\quad -\dfrac { \tau }{ tan\theta } \$\$\quad U\quad =-\dfrac { 4\sqrt { 3 } }{ \sqrt { 3 } } =-4Nm\$

Q: An electric dipole of length 2 cm, when placed with its axis making an angle of \$60^o\$ with a uniform electric field, experiences a torque of \$8\sqrt 3\$ Nm. Calculate the potential energy of the dipole, if it has a charge of \$\pm\$ 4nC.

Torque, \$\tau = PEsin \ \theta = 2Qlsin \ \theta \$ Here \$l\$ is the length of the dipole, \$Q\$ is the charge and \$E\$ is the electric field. Potential energy, \$U=-PEcos \ \theta=-2Ql cos \ \theta\$ \$\dfrac { \tau }{ U } =\dfrac { 2Qlsin\theta }{ -2Qlcos\theta } \ \ \ \ \ (where \ P=qd)\\ \ \ \ =-tan\theta \\ \dfrac { \tau }{ U } =-tan\theta \\ U=\dfrac { \tau }{ -tan\ 60^0 } =-\dfrac { 8\sqrt { 3 } }{ \sqrt { 3 } } =-8 \ joule\$

Q: An electric dipole of diploe moment \$\overrightarrow { p } \$ placed in uniform electric field \$\overrightarrow { E } \$ has maximum potential energy when angle between \$\overrightarrow { p } \$ and \$\overrightarrow { E } \$

\$\textbf{Step 1: Electric Potential energy of dipole}:\$The potential energy of the dipole is given by the equation,\$U= -\vec{p}.\vec{E}= -pE\cos\theta\$Here, \$\theta\$ is the angle between the dipole moment \$(\vec{p})\$ and the electric field\$(\vec{E})\$ as shown in figure.\$\textbf{Step 2: Condition for maximum Potential energy of dipole}:\$\$U\$ is maximum when \$\cos\theta= -1\$ and \$U=+pE\$\$\Rightarrow \theta= \pi\$Hence, Option(C) is correct.⃗

Q: A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. If q was given a speed 2v, the closest distance of approach would be:

As initially q is at infinite distance from Q, thus \$P.E_i=0\$For case 1: the closest distance given is r and velocity of q is v initially.Applying conservation of energy, \$P.E_i+K.E_i=P.E_f+K.E_f\$\$0+ \dfrac{1}{2}mv^2= \dfrac{KQq}{r}+ 0\$\$\implies \$\$ \dfrac{1}{2}mv^2= \dfrac{KQq}{r}\$ ................(1)For case 1: Let the closest distance be r' and velocity of q is 2v initially.Applying conservation of energy, \$P.E_i+K.E_i=P.E_f+K.E_f\$\$0+ \dfrac{1}{2}m(2v)^2= \dfrac{KQq}{r'}+ 0\$\$\implies \$\$ \dfrac{1}{2}m(2v)^2= \dfrac{KQq}{r'}\$ ................(2)Solving (1) and (2), \$r'=\dfrac{r}{4}\$

Q: An electric dipole of moment 'p' is placed in an electric field of intensity 'E'. The dipole acquires a position such that the axis of the dipole makes an angle 0 with the direction of the field. Assuming that the potential energy of the dipole to be zero when \$\theta = 90^0\$, the torque and the potential energy of the dipole will respectively be:

Potential energy of the dipole\$U=-\int { T } dB\$ \$=-\int _{ \pi /2 }^{ \theta }{ PE\sin\theta } d\theta \$ \$=PE\left( \cos\theta -0 \right) =-PE\cos\theta \$and potential energy \$=PE\sin\theta \$.

Q: An electric dipole when placed in a uniform electric field \$E\$ will have a minimum potential energy if the dipole moment makes the following angle with \$E\$

\$\textbf{Step 1: Potential energy of dipole }\$ Potential energy of dipole \$\vec p\$ in a uniform electric field \$\vec E\$ is given by: \${ U }=-\vec p . \vec E=-pE\cos { \theta } \$ \$....(1)\$Where \$\theta \$ is angle between electric field and dipole moment vectors.\$\textbf{Step 2: Condition of Minimum Potential Energy }\$ In Equation \$(1)\$, \$p\$ and \$E\$ are constant. \$U_{min} = -pE\$So for Minimum Potential Energy, \$cos\theta\$ should be Maximum \$= 1\$ \$\therefore\ \ \ [cos\theta]_{max} = 1\$\$\Rightarrow\ \ \theta = 0^o\$Hence, Option C is correct.\$\textbf{Note:}\ \theta = 0^o\$ means that dipole is parallel to the Electric Field as shown in Figure 2. And Minimum Potential energy implies that this position is stable equilibrium

Q: A particle of mass \$10^{-3}kg\$ and charge \$5\mu C\$ is thrown at a speed \$20\ m\ s^{-1}\$ against a uniform electric field of strength \$2\times 10^{5}N\ C^{-1}\$. How much distance will it travel before coming to rest momentarily?

Force on particle\$=F=qE\$ in opposite direction of motionAnd, \$F=ma=qE\$\$\implies a=\dfrac{qE}{m}=\dfrac{5\times 10^{-6}\times 2\times 10^5}{10^{-3}}\$\$\implies a=10^3m/s^2\$ and this acceleration is negative since particle is thrown against force.And final velocity is \$v=0\$Using \$v^2-u^2=2as\$\$\implies 0-20^2=-2\times 1000\times s\$\$\implies s=0.2m\$Answer-(D)

Question 1

Derive an expression for the torque acting on an electric dipole kept in a uniform electric field and hence define electric dipole moment.

Accepted Answer

Electric dipole moment is the measure of separation of positive and negative charges.  It is the measure of overall polarity.  Its SI unit is Coulomb-meter.

Question 2

Derive the formula for the electric potential energy of an electric dipole in a uniform electric field.

Accepted Answer

Potential energy of dipole in uniform electric field.$\vec {\tau}=\vec p\times \vec E$work done in rotating dipole by small angle$dw=\tau d\theta$$=-p\varepsilon \sin \theta d\theta$The change in potential energy $dU=-dW=pE\sin \theta d\theta$If $\theta \rightarrow 90^o$ to $\theta^o$$U(\theta) -U(90^o)=\displaystyle \int_{90^o}^{0^o}{P\varepsilon \sin \theta d\theta}$$=P\varepsilon [-\cos \theta]_{90}^{0}$$=-\vec P.\vec E$So$\boxed{U(\theta)=-\vec P. \vec E}$

Question 3

An electric dipole of length 4cm when placed with its axis making an angle of ${60^ \circ }$ with a uniform electric field, experience a torque of $4\sqrt 3 N - m$. Calculate the potential energy of the dipole of it has charge $ \pm 8nC$.

Accepted Answer

Dipole moment(P) = q.2L$P=8\times { 10 }^{ -9 }\times 2\times 4\times { 10 }^{ -2 }\ \quad =64\times { 10 }^{ -11 }Cm$Torque on the dipole in uniform electric field$\tau =PEsin\theta \ E=\dfrac { \tau  }{ Psin\theta  } $Potential energy of the dipole is $U=-PEcos\theta $on substituting the value , we get the value of E and U$U=-P\dfrac { \tau  }{ Psin\theta  } cos\theta \ \quad \quad =\quad -\dfrac { \tau  }{ tan\theta  } $$\quad U\quad =-\dfrac { 4\sqrt { 3 }  }{ \sqrt { 3 }  } =-4Nm$

Question 4

An electric dipole of length 2 cm, when placed with its axis making an angle of $60^o$ with a uniform electric field, experiences a torque of $8\sqrt 3$ Nm. Calculate the potential energy of the dipole, if it has a charge of $\pm$ 4nC.

Accepted Answer

&#10;&#10;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#10;&#10;&#10;&#10;Torque, $\tau =&#10;PEsin \ \theta = 2Qlsin \ \theta $&#10;Here $l$ is the&#10;length of the dipole, $Q$ is the charge and $E$ is the electric&#10;field.&#10;Potential energy,&#10;$U=-PEcos \ \theta=-2Ql cos \ \theta$&#10;$\dfrac { \tau  }{&#10;U } =\dfrac { 2Qlsin\theta  }{ -2Qlcos\theta  }  \  &#10;\   \   \   \  (where \  P=qd)\  \  &#10;\   \  =-tan\theta \ \dfrac { \tau  }{ U }&#10;=-tan\theta \ U=\dfrac { \tau  }{ -tan\ 60^0  } =-\dfrac {&#10;8\sqrt { 3 }  }{ \sqrt { 3 }  } =-8 \  joule$&#10;&#10;

Question 5

A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge

Accepted Answer

Equipotential surface is always perpendicular to the direction of electric field. Positive charge experiences the force in the direction of electric field.When a positive charge is released from rest in uniform electric field, its velocity increases in the direction of electric field. So K.E. increases, and the P.E. decreases due to law of conservation of energy.So P.E. of positively charged particle decreases because speed of charged particle moves in the direction of field due to force $q \bar{E}$

Question 6

An electric dipole of diploe moment $\overrightarrow { p } $ placed in uniform electric field $\overrightarrow { E } $ has maximum potential energy when angle between $\overrightarrow { p } $ and $\overrightarrow { E } $

Accepted Answer

$\textbf{Step 1: Electric Potential energy of dipole}:$The potential energy of the dipole is given by the equation,$U= -\vec{p}.\vec{E}= -pE\cos\theta$Here, $\theta$ is the angle between the dipole moment $(\vec{p})$  and the electric field$(\vec{E})$ as shown in figure.$\textbf{Step 2: Condition for maximum Potential energy of dipole}:$$U$ is maximum when $\cos\theta= -1$ and $U=+pE$$\Rightarrow \theta= \pi$Hence, Option(C) is correct.&#8407;

Question 7

A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. If q was given a speed 2v, the closest  distance of approach would be:

Accepted Answer

As initially q is at infinite distance from Q, thus $P.E_i=0$For case 1:  the closest distance given is r and velocity of q is v initially.Applying conservation of energy,  $P.E_i+K.E_i=P.E_f+K.E_f$$0+ \dfrac{1}{2}mv^2= \dfrac{KQq}{r}+ 0$$\implies $$ \dfrac{1}{2}mv^2= \dfrac{KQq}{r}$     ................(1)For case 1:  Let the closest distance be r' and velocity of q is 2v initially.Applying conservation of energy,  $P.E_i+K.E_i=P.E_f+K.E_f$$0+ \dfrac{1}{2}m(2v)^2= \dfrac{KQq}{r'}+ 0$$\implies $$ \dfrac{1}{2}m(2v)^2= \dfrac{KQq}{r'}$     ................(2)Solving (1) and (2),   $r'=\dfrac{r}{4}$

Question 8

An electric dipole of moment 'p' is placed in an electric field of intensity 'E'. The dipole acquires a position such that the axis of the dipole makes an angle 0 with the direction of the field. Assuming that the potential energy of the dipole to be zero when $\theta = 90^0$, the torque and the potential energy of the dipole will respectively be:

Accepted Answer

Potential energy of the dipole$U=-\int { T } dB$    $=-\int _{ \pi /2 }^{ \theta  }{ PE\sin\theta  } d\theta $    $=PE\left( \cos\theta -0 \right) =-PE\cos\theta $and potential energy $=PE\sin\theta $.

Question 9

An electric dipole when placed in a uniform electric field $E$ will have a minimum potential energy if the dipole moment makes the following angle with $E$

Accepted Answer

$\textbf{Step 1: Potential energy of dipole }$ Potential energy of dipole $\vec p$ in a uniform electric field $\vec E$ is given by: ${ U }=-\vec p . \vec E=-pE\cos { \theta  } $                $....(1)$Where $\theta $ is angle between electric field and dipole moment vectors.$\textbf{Step 2: Condition of Minimum Potential Energy }$  In Equation $(1)$, $p$ and $E$ are constant. $U_{min} = -pE$So for Minimum Potential Energy, $cos\theta$ should be Maximum $= 1$ $\therefore\ \ \ [cos\theta]_{max} = 1$$\Rightarrow\ \ \theta = 0^o$Hence, Option C is correct.$\textbf{Note:}\  \theta = 0^o$ means that dipole is parallel to the Electric Field as shown in Figure 2. And Minimum Potential energy implies that this position is stable equilibrium

Question 10

A particle of mass $10^{-3}kg$ and charge $5\mu C$ is thrown at a speed $20\ m\ s^{-1}$ against a uniform electric field of strength $2\times 10^{5}N\ C^{-1}$. How much distance will it travel before coming to rest momentarily?

Accepted Answer

Force on particle$=F=qE$ in opposite direction of motionAnd, $F=ma=qE$$\implies a=\dfrac{qE}{m}=\dfrac{5\times 10^{-6}\times 2\times 10^5}{10^{-3}}$$\implies a=10^3m/s^2$ and this acceleration is negative since particle is thrown against force.And final velocity is $v=0$Using  $v^2-u^2=2as$$\implies 0-20^2=-2\times 1000\times s$$\implies s=0.2m$Answer-(D)

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