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: A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of (i) capacitance.(ii) potential difference between the plates.(iii) electric field between the plates.(iv) the energy stored in the capacitor.

\$(i)\$ The capacitance increases as the dielectric constant \$K > 1\$. \$(ii)\$ Potential difference \$V =\dfrac{Q}{C}\$. As \$C\$ increases and \$Q\$ remains the same since the battery is disconnected, the p.d. between the plates decreases.\$(iii)\$ Electric field \$E =\dfrac{V}{d}\$ where \$V\$ is the p.d. and \$d\$ the separation between the plates. As \$V\$ decreases and d remains the same, electric field also decreases.\$(iv)\$ Energy stored in a capacitor \$U =\dfrac{1}{2}\dfrac{{Q}^{2}}{C}\$. As \$Q\$ is constant and \$C\$ increases, \$U\$ decreases.

Q: (a) Deduce the expression for the potential energy of a system of two charges \$q_1\$ and \$q_2\$ located \$\overrightarrow{r_1}\$ and \$\overrightarrow{r_2}\$, respectively, in an external electric field.(b) Three point charges, + Q + 2Q and - 3Q are placed at the vertices of an equilateral triangle ABC of side l. If these charges are displaced to the mid-point \$A_1, B_1\$ and \$C_1\$, respectively, find the amount of the work done in shifting the charges to the new locations.

\$\left( a \right) \rightarrow (i)\$ work done to bring a charge q in the electric field at a distance \${ r }_{ 1 }={ q }_{ 1 }V\left( { r }_{ 1 } \right) \$ \$(ii)\$ work done to bring a charge q in the electric field at a distance \${ r }_{ 2 }={ q }_{ 2 }V\left( { r }_{ 2 } \right) \$ \$(iii)\$ work done on \${ q }_{ 2 }\$ to move it against the force of \${ q }_{ 1 }=\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { r }_{ 12 } } \$\$\therefore \$ The potential energy of the system \$={ q }_{ 1 }V\left( { r }_{ 1 } \right) +{ q }_{ 2 }\left( V \right) \left( { r }_{ 2 } \right) +\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { r }_{ 12 } } \$\$(b)\$ work done in moving charges \$=\$ Potential of the initial position - Potential of the final position.Now, Potential of the initial system \$={ V }_{ 12 }+{ V }_{ 23 }+{ V }_{ 31 }={ 2KQ }^{ 2 }+\left( { -3KQ }^{ 2 } \right) +\left( -6K{ Q }^{ 2 } \right) =-7K{ Q }^{ 2 }J\$potential energy after moving to position, \${ A }_{ 1 },{ B }_{ 1 }\& { C }_{ 1 }\$\$=\dfrac { K\left( 2Q \right) \left( Q \right) }{ 1/2 } +\dfrac { K\left( Q \right) \left( -3Q \right) }{ 1/2 } +\dfrac { K\left( -3Q \right) \left( 2Q \right) }{ 1/2 } \$\$=4{ KQ }^{ 2 }-6{ KQ }^{ 2 }-12{ KQ }^{ 2 }=-14K{ Q }^{ 2 }J\$\$\therefore \$ work done \$=-7{ Q }^{ 2 }K-\left( -14K{ Q }^{ 2 } \right) =7K{ Q }^{ 2 }J\$

Question 1

Derive the expression for capacitance of parallel plate capacitor.

Accepted Answer

Let the two plates are kept parallel to each other separated be a distance $d$ and cross-sectional area of each plate is $A$.  Electric field by a single thin plate $E' = \dfrac{\sigma}{2\epsilon_o}$Total electric field between the plates $E = \dfrac{\sigma}{2\epsilon_o} +\dfrac{\sigma}{2\epsilon_o}$Or $E = \dfrac{\sigma}{\epsilon_o}$Or $E = \dfrac{Q}{A\epsilon_o}$Potential difference between the plates $V = Ed$$\implies$ $V = \dfrac{Qd}{A\epsilon_o}$Capacitance  $C = \dfrac{Q}{V}$Thus we get capacitance of parallel plate capacitor  $C  =\dfrac{A\epsilon_o}{d}$

Question 2

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 3

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 4

Two charges $ 5 \times 10^{-8}\ C$ and $-3\times 10^{-8}\ C$ are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Accepted Answer

Let p be the point of zero potential be p at distance r from the charge 1..$d=16\ cm$Electric potential, $v = q_1/4\pi\epsilon_or+ q_2/4\pi\epsilon_o(d-r)$for $v=0$, $\displaystyle \frac{q_1}{r}=-\frac{q_2}{d-r}$$\Rightarrow \displaystyle \frac{5\times 10^{-8}}{r}=\frac{-3\times 10^{-8}}{0.16-r}$$r=10\ cm$If, point p is outside,$v = q_1/4\pi\epsilon_or+ q_2/4\pi\epsilon_o(r-d)$so here, $r=40\ cm$

Question 5

Derive an expression for energy stored in a capacitor. In which form energy is stored?

Accepted Answer

Let us consider a capacitor of capacitance C and potential difference V between the plates.Let the charge on one plate be +q and -q on the other.Suppose the capacitor is being charged gradually.Now,at any stage the charge on capacitor is q.Therefore, the potential difference = $ \frac{q}{C} $Small amount of work doe in giving n additional charge dq to the capacitor isdW= $\dfrac{q}{C}*dq$Total work done in giving a charge Q to the capacitor  isW=$\int dW$W= $ \int_{0}^{Q} \frac{Q}{C}dq $ W = $\dfrac{Q^{2}}{C}$Energy = EE= $\dfrac{Q^2}{2C}$ = $\dfrac {CV^2}{2}$ = $\dfrac{QV}{2}$The energy is stored in the form of potential energy.

Question 6

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of (i) capacitance.(ii) potential difference between the plates.(iii) electric field between the plates.(iv) the energy stored in the capacitor.

Accepted Answer

$(i)$ The capacitance increases as the dielectric constant $K > 1$. $(ii)$ Potential difference $V =\dfrac{Q}{C}$.  As $C$ increases and $Q$ remains the same since the battery is disconnected, the p.d. between the plates decreases.$(iii)$ Electric field $E =\dfrac{V}{d}$ where $V$ is the p.d. and $d$ the separation between the plates. As $V$ decreases and d remains the same, electric field also decreases.$(iv)$ Energy stored in a capacitor $U =\dfrac{1}{2}\dfrac{{Q}^{2}}{C}$. As $Q$ is constant and $C$ increases, $U$ decreases.

Question 7

Derive an expression for the electric potential due to a point charge :

Accepted Answer

Consider the electric potential due to a point charge q, As we move from point A, at distance rA from the charge q, to point B, at distance rB from the charge q, the change in electric potential is$\Delta V_{BA}= V_B-V_A=- \int_A^B E. ds$$E.ds=[k\dfrac{q}{r^2}]\hat{r}.ds$$\hat{r}.ds=dr$Only the radial distance r determines the work done or the potential. We can move through any angle we like and, as long as the radial distance remains constant, no work is done or there is no change in the electric potential.$\Delta V_{BA}= V_B-V_A = - \int_{r_A}^{r_B}Edr=-\int_{r_A}^{r_B}[k\dfrac{q}{r^2}dr]$$\Delta V_{BA}= V_B-V_A=-kq[(-1)r^{-1}]_{r_A} ^{r_B}$$\Delta V_{BA}= V_B-V_A= kq [\dfrac{1}{r_B}-\dfrac{1}{r_B}]$This is the change in electric potential due to a point charge as we move from rA to rB.We could ask about the change in electric potential energy as we move a charge q' from radius rA to rB due to a point charge q$\Delta U_{BA}= kq'q[\dfrac{1}{r_B}-\dfrac{1}{r_A}]$As with gravitational potential energy, it is more convenient -- and, therefore, useful -- to talk about the electric potential energy or the electric potential relative to some reference point. We will choose that reference point to be infinity. That is,$r_A=\infty$That means we can then write the electric potential at some radius r as $V=kq\dfrac{1}{r}$

Question 8

Two charges $q_1$ and $q_2$ are placed 30 cm apart, as shown in the figure. A third charge $q_3$ is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is $\frac {q_3}{4\pi \epsilon_0}k$, where k is

Accepted Answer

We know that potential energy of discrete system of charges is given by$U=\frac{1}{4\pi\epsilon_0}\left[\frac{q_1q_2}{r_{12}}+\frac{q_2q_3}{r_{23}}+\frac{q_1q_3}{r_{13}}\right]$Here, $AB=0.3  m, AC=AD=0.4  m,   BC=\sqrt{0.3^2+0.4^2}=0.5  m,  BD=AD-AB=0.4-0.3=0.1  m$Initial potential energy of the system is $\displaystyle U_i=\frac{1}{4\pi\epsilon_0}\left[\frac{q_1q_2}{AB}+\frac{q_2q_3}{BC}+\frac{q_1q_3}{AC}\right]$$\displaystyle =\frac{1}{4\pi\epsilon_0}\left[\frac{q_1q_2}{0.3}+\frac{q_2q_3}{0.5}+\frac{q_1q_3}{0.4}\right]$Final potential energy of the system is $\displaystyle U_f=\frac{1}{4\pi\epsilon_0}\left[\frac{q_1q_2}{AB}+\frac{q_2q_3}{BD}+\frac{q_1q_3}{AD}\right]$$\displaystyle =\frac{1}{4\pi\epsilon_0}\left[\frac{q_1q_2}{0.3}+\frac{q_2q_3}{0.1}+\frac{q_1q_3}{0.4}\right]$The change in the potential energy of the system is$\displaystyle U=U_f-U_i=\frac{1}{4\pi\epsilon_0}\left[\frac{q_2q_3}{0.1}+\frac{q_1q_3}{0.4}-\frac{q_2q_3}{0.5}-\frac{q_1q_3}{0.4}\right]=\frac{1}{4\pi\epsilon_0}{8q_2q_3}=\frac{q_3}{4\pi\epsilon_0} 8q_2$

Question 9

(a) Deduce the expression for the potential energy of a system of two charges $q_1$ and $q_2$ located $\overrightarrow{r_1}$ and $\overrightarrow{r_2}$, respectively, in an external electric field.(b) Three point charges, + Q + 2Q and - 3Q are placed at the vertices of an equilateral triangle ABC of side l. If these charges are displaced to the mid-point $A_1, B_1$ and $C_1$, respectively, find the amount of the work done in shifting the charges to the new locations.

Accepted Answer

$\left( a \right) \rightarrow (i)$ work done to bring a charge q in the electric field at a distance ${ r }_{ 1 }={ q }_{ 1 }V\left( { r }_{ 1 } \right) $            $(ii)$ work done to bring a charge q in the electric field at a distance ${ r }_{ 2 }={ q }_{ 2 }V\left( { r }_{ 2 } \right) $            $(iii)$ work done on ${ q }_{ 2 }$ to move it against the force of ${ q }_{ 1 }=\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { r }_{ 12 } } $$\therefore $ The potential energy of the system $={ q }_{ 1 }V\left( { r }_{ 1 } \right) +{ q }_{ 2 }\left( V \right) \left( { r }_{ 2 } \right) +\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { r }_{ 12 } } $$(b)$ work done in moving charges $=$ Potential of the initial position - Potential of the final position.Now, Potential of the initial system $={ V }_{ 12 }+{ V }_{ 23 }+{ V }_{ 31 }={ 2KQ }^{ 2 }+\left( { -3KQ }^{ 2 } \right) +\left( -6K{ Q }^{ 2 } \right) =-7K{ Q }^{ 2 }J$potential energy after moving to position, ${ A }_{ 1 },{ B }_{ 1 }\& { C }_{ 1 }$$=\dfrac { K\left( 2Q \right) \left( Q \right)  }{ 1/2 } +\dfrac { K\left( Q \right) \left( -3Q \right)  }{ 1/2 } +\dfrac { K\left( -3Q \right) \left( 2Q \right)  }{ 1/2 } $$=4{ KQ }^{ 2 }-6{ KQ }^{ 2 }-12{ KQ }^{ 2 }=-14K{ Q }^{ 2 }J$$\therefore $ work done $=-7{ Q }^{ 2 }K-\left( -14K{ Q }^{ 2 } \right) =7K{ Q }^{ 2 }J$

Question 10

A parallel plate capacitor is charged an then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then

Accepted Answer

When the capacitor is charged and then disconnected, the charge across the plates remains a constant, even when the plates are separated.The capacitance is related to the structure of parallel plate capacitor as $\dfrac{A\epsilon_0}{d}$.When the plates are separated, the distance between the plates $d$ increases and hence the capacitance decreases.Hence from $Q=CV$, the voltage across the capacitor increases in order to keep the charge constant.